Thermo Review Solutions

Here are the solutions to the packet I handed out on Friday:

http://groups.google.com/group/bronx-ap-physics/web/Thermo+Review+Problem+Solutions.pdf

As requested, I will have a couple problems for you to play around with, but I expect most of what we do will come from you!

Bronx Tournament is ON!

The Bronx FLL Tournament is going on as scheduled today, Saturday, December 20, 2008.

Please leave yourself extra time to get to the event safely.

See you all there!

Thermo-cycling in the Snow!

Handout answers:

1
a) T = 481 K
b) Wnet = +4000 J
c) For a full cycle, delta_U = 0. This means that for the entire cycle, Q = -W = -4000 J.
d) For any cycle, delta_U is zero, so it stays the same.

2.
a) AB,CD - isobaric, BC, DA - isochoric
b) 1.202 x 10^-4 moles (use P,V, and T at state D.)
c) Wab = -1.0 J, Wad = 0.5 J, all others zero
d) net work for cycle is the area inside = -0.5 J
e) Tb = 800 K, so delta_U = 0.750 J
f) P = Wnet/delta_t = 0.5 J/1.2 s = .417 W

3. b, c, a (remember that in c and a, the work is negative, and C has a work that is LESS negative.)

In This House, we obey the laws of Thermodynamics!

Handout solutions:

1.
a) Tf = 400 K
b) The ratio of average kinetic energies becomes the ratio of the temperatures Tb/Ta = 3/4 since the Boltzmann constant divides out. (KE average = 3/2 kb*T)
c) W = -P*delta_V, but we have to find the pressure first. P = nRT/V = 8315 Pa.
W = -831.5 J
d) using 1st law, Q = delta_U - W = 1.5*(1 mole)(8.315 J/mol*K)(400 K - 300 K) - (-831.5 J) = 2,079 J

2.
a) 0.3 kg/(.004 kg/mol) = 75 moles
b) using 1st law, delta_U = Q + W. We can apply this to the entire process. During the first part, we can assume W = 0 and Q = +3100J. During the second, Q = 0, and W = +1000 J. Thus for the entire process, Q = 3100 J, and W = 1000 J. This means that delta_U = 4100 J.
c) Since delta_U = 4100 J = 1.5nRdelta_T, delta_T = 4.38 K. The final temperature is then 254.4 K.

3. It might help to draw a PV diagram on this problem. There are two segments, and three states to consider.

The initial temperature is 273 K, initial pressure = 101,300 Pa. We can use PV=nRT to find that the initial volume is 2.7 x 10^-2 m^3.

At state 2, the volume is 2.7 x 10^-2 m^3 (const. volume) and P = 303900 Pa. The temperature is therefore 3*273 K =m 819 K.

At state 3, the volume is 5.4 x 10^-2 m^3, and P = 303900 Pa. The temperature at state 3 is therefore 2*819 K = 1638 K.

For this whole process, the work is -(303,900 Pa)(5.4 - 2.7)*10^-3 m^3 = -820.5 J.

The change of internal energy is 3/2*n*R(delta_T) = 17030 J.

From the 1st Law, this means that Q = 17030 J - (-820.5 J) = 17850 J.

Book Problems:

p. 440
32. H = k*A*delta_T/L = (0.84 J/s · m · C°)(3.0 m^2)[15°C - (- 5°C)]/(3.2 x 10-3 m) = 1.6 x 10⁴ W.

p. 471 - 472:
Q3. It is enough information since the process is isothermal - delta_U = 0, so Q = -W.

Q6. In an adiabatic compression, Q = 0, and the work is positive. By the first law, this means delta_U is positive, which means delta_T is positive, an indication that temperature increases.

Problems:
6.
a) Volume doesn't change, so W = 0.
b)Q = -265 kJ, W = 0, so delta_U = -265,000 J.

7.
a) adiabatic means Q = 0.
b) Q = 0, W = 1350 J, so delta_U = 1350 J.
c) since delta_U is positive, the temperature must increase.

8.
a) Work is done only during the constant pressure process, and is equal to -P*delta_V = -1300 J.
b) Q = delta_U - W but since the final temperature is the same as the original, delta_U = 0.

Q = 0 - (-1300 J) = 1300 J.

9.
a) W = -3500 J
b)since the ttemperatures are equal, delta_U = 0.
c) Q = delta_U - W = 0 - (-3500 J) = 3500 J, which represents a heat flow INTO the gas.

In the Thermodynamic Process of Decorating

Handout solutions:

5. We need to convert to Pascals and m^3: 4.0 L = 4 x 10^-3m^3 and 2.0 L = 4 x 10^-3m^3.

Since the initial gauge pressure is 3 atm, the absolute pressure is 4 atm = 405,200 Pa.

For a constant pressure process, W = -P*delta_V = -(405,200 Pa)(-2 x 10^-3m^3) = 810.4 J

b) Since it's a constant pressure process, the final pressure is also 405,200 Pa.

c) Using ideal gas law, T2 = T1*(V2/V1) = 149 K

6.
a) Using ideal gas law, 5P0*3V0 = nRT0 so T0 = 15P0V0/nR

b) Since temperature is constant, the change of internal energy is zero.

c) Using ideal gas law again, P2V2 = nRT2

We know that T2 = T0 = 15P0V0/nR so P2V2 = 15P0V0.

V2 = 6V0

P2 = 15P0V0/6V0 = 5/2P0

7.
a) Ideal gas law - n = PV/nT = (7*101,300Pa)(10 x 10^-3m^3)/(8.315.J/Mol*K * 500 K) = 1.706 moles

b) The container volume is constant, so the work done is zero.

c) T2 = T1*P2/P1 = 214.3 K

d) THe graph should be a vertical line at V = 10 x 10^-3 m^3 from P = 7 atm to 3 atm (or the equivalent pressures in Pa.)

Book HW:

3,4,10abc, 53a,c

3.

4.

10
a) and c) PV diagram:

b) W = -2700 J, found since the process is isobaric, using W = -P*delta_V

53.
a) W = -2.2 x 10⁵ J

c)

In Theory, We Trust - Kinetic Theory Solutions

A bonus assignment for tonight - if you check your answers, please log in and write a comment (go to the bottom of the post for the link) stating the name of your favorite ideal gas. If you don't have one, choose Helium.

50. The gas starts at 0 degrees C, or 273 K. Since v = (3RT/m)^1/2, in order to double v, the temperature must be multiplied by 4. Thus the temperature must be 1092 K.

52. Following the similar logic as with #50, If the pressure doubles, and volume stays the same, the ideal gas equation says that the temperature will also double. This will make the v_rms increase by a factor of the square root of 2.

55. You could get to the solution of this equation by looking at the derivation we did today. We ended up with the fact that P = (1/3)m*N*v^3/(V).

The quantity m*N is the total mass of the gas contained in the container, as it's the mass of a single molecule times the number of molecules. The quantity M*N/V must then bethe density of the gas.

Also from the derivation, the v in our expression is the mean square speed of the molecules in the gas.

Thus P = (1/3)*density*v^2

There are other ways to derive it using the other formulas we came up with today, so if your method is different, that is fine.

80. Since you know the temperature, you can directly calculate the v_RMS = 183.5 m/s.

Using the result of question 5, we have a relationship between the density, v_RMS, and the pressure.

The density will be one atom/cm^3 = 1 amu * 1.66 x 10^-27 kg*(10⁶cm^3/m^3) = 1.66 x 10^-21 kg/m^3.

Substituting, the pressure is 1.9 x ^-17 Pa = 1.84 x ^-22 atm

81. When 10 meters below the surface, the absolute pressure is Patm + rho*g*h = 199,300 Pa.
At the surface, the pressure is 101,300 Pa. You also know the initial volume underneath the water, as well as the fact that the moles of air in the diver's lungs are constant as he/she rises to the surface.

If we assume the rise happens, quickly, we can also say that the temperature is approximately constant.

Thus P1V1/P2V2 = nRT1/nRT2. The entire right side of the equation divides out to 1.

Thus V2 = P1V1/P2 = 10.8 L. I personally would NOT want my lungs stretched to three times their original volume, nor would I expect it to be medically advisable for ANYONE to have this done to them.

This is one of the reasons why it's important for scuba divers to take their time rising to the surface after a dive. One of them is that it causes a condition called 'the bends', where dissolved gas comes out of solution in the blood stream. This is a bad thing. Another is that, without releasing any gas in the lungs during the resurfacing process, the lungs will expand rapidly during the rise, possibly causing tearing in the tissue.

Expand your Mind, and Think of an Ideal Gas - Solutions

p.413, 12, 13, 34, 35, 38, 40 (What is V2/V1?)

12. Notice that you can do this problem without converting to meters, since you end up dividing delta_L by L0, which divides out the units anyway.

You should be able to find that delta_T = delta_L/(alpha*L0) = -89 degrees. (Remember change is final minus initial!) Adding this to the initial temperature of 20 degrees Celsius gives the final temperature of -69 degrees.

13. The key here is multiplying the final lengths of the plate after expansion.

(L + dL)(W + dW) = LW + W*dL + L*dW + dL*dW

As I mentioned in class, the last term is a higher order term, and can be neglected.

If we then substitute using the thermal expansion equation for dW = alpha*W*dT and dL = alpha*L*dT, we get that the final area is LW + 2*alpha*LW*dT.

The question asks for the change in area, so we just have to subtract the initial area LW to get what the text says should be the answer, 2*alpha*L*W*dT.

34. The first task is to determine the number of moles of the gas. Since there are 18.5 kg (18,500 g) of Nitrogen, and the molar weight of Nitrogen gas is 28 g/mol, we can divide to get that there are 661 moles of N2 gas in the tank.

We now have enough information to solve for V = nRT/P (remember that standard temperature and pressure means 273 K, 1.013 x 10⁵ Pa)

V = 14.8 m^3.

For part b, the tank has not changed, so it has the same volume as you found for the first part. We do know there is a new number of moles in the tank (661 + additional 536 moles).

Since we are looking for pressure, we again must solve for P in the ideal gas equation.

P = nRT/V = 1.84 x 10⁵ Pa

35. Remember we have to use absolute pressure, so the initial pressure in the container is (1 atm + .35 atm)*(1.013 x 10⁵ Pa/atm) = 1.37 x 10⁵ Pa

Solving, V = nRT/P = 0.439 m^3

b) Here we are changing states for the gas, so we need to use the division trick from class.

P1V1 = nRT1 and P2V2 = nRT2 , which can be combined to give
(P2/P1)(V2/V1) = T2/T1

Solving, T2 = 210 K = -63 degrees Celsius.

38. IN this case, since it is a change of state problem, we do NOT have to convert the pressure to Pascals before doing anything. We can use the same division method as before.

(P2/P1)(V2/V1) = T2/T1

P2 = (T2/T1)(V1/V2)*P1

P2 = (323.2 K/291.2 K)*(55.0 L/48.8 L)* 1 atm = 3.03 atm

40. Again, using the same method:
(P2/P1)(V2/V1) = T2/T1

(V2/V1) = (T2/T1)*(P1/P2) = 1.4

Fluid Solutions

Click on each picture to make it large.

Bernoulli and Friends - It's a Family Program!

Q 18,19 P37, 39, 41, 43, 72 (pressure head is the source of the water), 74

We will have a quiz Tuesday on fluid mechanics, so tomorrow (Monday) will be a day of practicing problems and deciding which method to use to solve parts of problems.


Questions:
18. When the streamlines of the wind curve over the roof, this creates a low pressure area above the roof, while the pressure below the roof remains high. This difference in pressure creates an upward force that pushes the roof off the house.

19. The increase in speed of the air passing between the sheets causes a lower static pressure between them. Since the air pressure on the outside of the sheets is the same, the outside pressure pushes into the space between where the pressure is lower.

Problems:

37. Using Bernoulli, with 1 at the ground and 2 at the top of the water flow, we can assume that at the MINIMUM pressure, the fluid is not moving.

P1 + 0 + 0 = Patm + 0 + rho*g*h

P1 - Patm = 1.176 x 10⁵ Pa

39. The pressure is atmospheric underneath the roof. We can take 1 to be at a point far from the roof (where it is still), and 2 to be in the wind above the roof:

Patm + 0 + 0 = P2 + 1/2*rho*v^2 + 0

The net pressure difference will be the gauge pressure of P2, or P2 - Patm

P2 - Patm= - 1/2*rho*v^2 = 580.5 Pa


Since this is pressure, the force is this pressure times the area of the roof:
F = 580.5 Pa * 240 M^2 = 1.39 x 10⁵ N

41.
Consider a streamline that goes from the outside of the hurricane (where P = Patm and the air is still) to the center. Remember to convert 300 km/h to 83.3 m/s.

Patm + 0 + 0 = P2 + .5*rho*v^2 + 0

P2 = 9.68 x 10⁴ Pa

43. We have to use continuity to find the speed of the water in the pipe when the pipes have smaller diameter. Since it is a liquid, we can assume the flow is incompressible.

A1v1 = A2v2, so v2 = A1/A2*v1 = 2.22 m/s

The gauge pressure at the pump is 3.8 atm gauge, which means it is 4.8 atm = 4.86 x 10⁵ Pa. The question asks for the static pressure at the top of the building.

Bernoulli:

P_pump + .5*rho*v1^2 + 0 = P_top + .5*rho*v2^2 + rho*g*h

P_top = 2.88 x 10⁵ Pa absolute, or 1.87 x 10⁵ Pa gauge.

72. The pressure head is the source of the water coming out of the shower, a tank on the roof perhaps. We can assume that the water has no velocity initially in the tank, and that the initial pressure is atmospheric.

Patm + 0 + rho*g*H = Patm + .5*rho*v2^2 + 0
H = 2.64 m

74. If the raft holds the maximum number of people, the logs are completely submerged in the water, though not yet sinking. This means the logs and people are in equilibrium. Note the factor of 10 since there are ten logs.

Fb - M_totalg = 0
10*rho_water*g*pi*R^2*L = (10*rho_wood*pi*R^2*L + M_people)*g

Since we know the specific gravity of wood, we can solve for the mass of the people.

M_people = 2087 kg = N*70 kg
N = 29.8 people

Thus there can be no more than 29 people on the raft.

We MUST Maintain Continuity - The Continuity Equation

Q10,13
P35,36,71

Question 10. For the barge to float with its top lower, the bottom of the barge must be at a greater depth. This means the buoyancy force must be greater. For this to happen and the barge still be in equilibrium, the mass of the boat must be greater. Therefore sand must be added to the barge.

Q 13. The buoyancy force depends only on g, the fluid density, and the submerged volume. The buoyancy force must therefore be the same whether it is submerged completely just below the surface, or deep below, since the submerged volume is the same in both cases.

Problems:
35. We are given the volume of the room, and are told that the air in the room is replaced in 10 minutes. This gives us the volumetric flow rate into/out of the room: Q = 0.345 m^3/s.

This is equal to the volumetric flow rate in, with A₁v₁ = Q.

Solving, v₁ = 3.80 m/s

36. The volume of the pool when it is filled is pi*(3.6 m)^2*(1.5 m) = 61.07 m^3

5/8 inch = .015875 m

Q = A₁v₁ = volume/delta_t

so delta_t = volume/(A₁v₁) = (61.07 m^3/(pi*(.015875 m)^2*.28 m/s) = 76.5 hours

71. Gauge pressure of -80 mm Hg means the atmospheric pressure is 680 mm Hg.

Using the conversion factor that 760 mm Hg = 1.013 x 10⁵ Pa

The pressure in the lungs is 9.064 x 10⁴ Pa

You should now imagine that the lungs are a closed container of gas on top of a column of water (like the problems we did on manometer day.)

Comparing the pressure at the bottom of the water column with the atmosphere:

P_lungs + rho_water*g*H = Patm

H = 1.088 m

The Problem is your Pressure Valve - I can get you a new one for $250

Handout problem solutions:

4.
a) The FBD has buoyancy force upwards, and weight (mg) downwards.
b) Fb = rho*g*Vsub where Vsub is the volume of the ball. Since the volume of a sphere is (4/3)*pi*R^3, and we know the density of water, we can calculate that the buoyancy force is
410.5 N 41.1 N.
c) Using the same method, but the density of the ball (800 kg/m^3) instead of water, we get that the mass is 33.5kg 3.35 kg.
d) Fnet in y = Fb - mg = ma, so a = (Fb - mg)/m = 2.454 m/s^2

5.
a) tray is in equilibrium, mg = rho*g*Vsub
Vsub = area of tray A*depth d, where A = L*w

mg = rho*g*L*w*d, m = 0.3 kg

b) Fbmax = rho*L*W*h = 23.52 N

c) If the tray is just on the edge of sinking, then Fbmax - (m_total)*g = 0
m_total = m_tray + m_weights

Solving, m_weights = 2.1 kg

Dividing by 0.005 kg per weight, this means there are 420 weights that could be added.

6.
a) Gauge pressure = 8 atm * (101.3 kPa/atm) = 8.103 x 10⁵ Pa

b) Gauge pressure = P - Patm = rho*g*h = 8.103 x 10⁵ Pa

c) F = P*A = 1.61 x 10⁶ N directed upwards

d) Same formula, F = P*A = 1.621 x 10⁶ N directed downwards

e) The difference between the two forces is 10,460 N directed downwards, so there needs to be an upwards force of 10,460 N to hold the window in place.

7.
a) The mercury will flow into the water since the pressure at the bottom is so much greater than the pressure of the water (P = P0 + rho*g*h, rho of mercury is much greater.)

b) When the mercury drops, we can say the distance it drops is H.

The height of the mercury column is then (1.0 m - H). Since the volume is conserved, the height of the water column is (1.0 m + H).

rho_Water*g*(1.0 m + H) = rho_Mercury*g*(1.0 m - H)

Solving for H, H = 1.863 m.

The Pressure is Getting to Me... - Manometers and Buoyancy

Solutions:

Sphere problem #3 on the handout from today:
a) When the sphere is not submerged, T - mg = 0 so m = T/g = 2 kg.

Once the sphere is submerged, T + Fb - mg = 0 so Fb = mg - T. Since Fb = rho*g*Vsub, V = (mg - T)/(rho*g) = .00153 m^3

b) density is m/V = 2 kg/(.00153 m^3) = 1307 kg/m^3

Book Problems:
22. Apparent mass comes from apparent weight. If you know the apparent weight (Fn) you can divide by g to get the apparent mass.

In this case, Fn = m_apparent * g = 60.6 N

Drawing a FBD of the rock at the bottom of the container, Fn + Fb - mg = 0. Fb = mg - Fn = rho*g*V. This enables us to get that the volume is .002016 m^3.

The density is therefore 8.2 kg/.002016 m^3 = 4067 kg/m^3


24. Answer: 3.0 x 10⁴ kg
Here's the FBD:

Since the balloon is in equilibrium, Fb - m_cargo*g - m_balloon*g - m_Helium*g = 0.

We can find the buoyancy force because we can find the volume of a sphere from the radius, and the density of air is 1.29 kg/m^3 (from p. 276). The same thing is true for knowing the mass of the Helium.

(I think in tutoring, I forgot about the mass of the Helium...)

The only unknown is the mass of the cargo, so you can solve!


26. Using the same method as in #9, we can find that the density of the metal is 8.94 x 10³ kg/m^3. From page 276, we can see that this is copper.

27. Same method, again. Density = 1.03 x 10³ kg/m^3

29. Finding the definition of specific gravity (SG) is a bit tricky in the chapter, which is why coming to tutoring made this a bit easier.

SG is the ratio of a density and the density of water. For example, if the SG = 0.79 for the alcohol, it means that the density is 790 kg/m^3.

Using this info, this is another problem that is along the lines of 9, 26, and 27. You should find that the density of the wood is 850 kg/m^3, which means that its specific gravity is 0.85.

30. This problem is a bit tricky because it's hard to picture what's happening. If we say that the iceberg is a cylinder, with cross section A, we can say that the part below the water is of depth D, and the part above the water is height H. The fraction the problem is asking for is H/(H+D).

Knowing the SG of both the iceberg and the ocean water, we know the densities of both. From a FBD, we can see that Fb - mg = 0.

Fb = rho_ocean*g*V_submerged = rho_ocean*g*A*(D)

mg = rho_ice*g*V = rho_ice*g*A*(H+D)

Setting these equal to each other, we get that D/(H+D) = rho_ocean/rho_ice = .895. The fraction above the water is then 1 - D/(H+D) = .105

Under Pressure - Fluid Properties

Remember, knowing the definitions and equations is CRUCIAL in fluid mechanics since the tough parts come from applying them

Question 3. The pressure is the same in each case, as well as the area, and this makes the force exerted on the bottom face of the container the same for all three. There is some downward pressure exerted on the diagonal sides in the second and third container. This makes up for the additional weight of water in the second and third container.

Problems:

9. This is straight plug-and-chug - the force is just P*A = 3.08 x 10⁵ N. The force on the bottom of the table is the same, as the atmospheric pressure is the same and the area is as well.


11. Make sure you convert the area into square meters! 200 cm^2 *(1 m/100 cm)^2 = .02 m^2

The total force is 4*P*A = 19,200 N = mg. Thus m = 1959 kg.


16. The key to this problem is in the hint - the pressure is the same at both a and b.

rho_oil*g*H1 + P₀ = rho_water*g*H2 + P₀

rho_oil = 654 kg/m^3

17. The total height of the water over the bottom of the hill is 5.0 m + 100 m sin 60 = 91.6 meters.

The total pressure at the bottom is given by the equation we derived today:

P = Patm + rho*g*h which is the absolute pressure at the bottom.

If we now subtract Patm from both sides, we get that the gauge pressure is rho*g*h = 8.98 x 10⁵ Pa.

18. This is similar to 17. P_absolute = Patm + rho*g*h so the gauge pressure is P_absolute - Patm = rho*g*h = 4.02 x 10⁵ Pa

19.
a) The volume of the water in the tube is pi*R^2*H = 3.39 x 10^-4 m^3. (Don't forget to convert to meters!) Since rho = M/V, M = rho*V = 0.339 kg.

b) The pressure on the inside surface of the barrel lid is given by Patm + rho*g*h. The pressure on the outside surface of the lid is Patm. Thus the net pressure will be the gauge pressure, rho*g*h. The force is therefore equal to the pressure times the area: rho*g*h*A = 1000 kg/m^3 * 9.8 m/s^2 * 12 m * pi*(.2 m)^2 = 1.48 x 10⁴ N

Midterm Reflection Form

Hi everyone,

Here is the link to the midterm reflection form. Fill it out and either email it to me (you will have to save it on your computer to do this) or print it out and give it to me tomorrow.

I am also posting your assignment now, in case you want to do it tonight before break.

Your assignment is to watch two lessons at http://www.montereyinstitute.org/courses/AP%20Physics%20B%20I/nroc%20prototype%20files/coursestartc.html. Click on Unit 2, and Chapter 7 for fluid mechanics.

You must watch the first lesson on hydrostatic pressure, but the second can be any of the four. You might also find that some of the lessons in Unit 1 help you understand what we have done so far in class - try them!

You should expect a quick quiz on some definitions when we return to see what you learned.

The Cable's Out - why else would there be so much STATIC?

Timed Problem Solutions: (Cannon on a spring problem)

a) Fn - mg = 0 so Fn = mg = (5200 kg)(9.8 m/s^2) = 50,960 N

b) momentum is conserved in the x-direction:
0 = m_projectilev_projectile*cos(45) - m_cannon*v

v = 3.536 m/s = final speed of cannon

c) .5*m*v^2 + 0 = 0 + .5*k*x^2
x = 1.768 m

d) F_spring = k*x = 20000 N/m * 1.768 m = 35,400 N

e) T = 2*pi*(m/k)^(1/2) = 1.99 s, so f = .5 Hz


Handout Solutions:
1.
a) I'm not drawing this diagram using ASCII characters.

b) F = 900 N

c) D = 5/9 L

2.
b) T sin(40)*(3L/4) - (800 N)*L - (600 N)*L/2 = 0

c) Px - T cos(40) = 0, so Px = 1748 N to the right ,

Tsin(40) - 800 N - 600 N - Py = 0, Py = 66.65 N directed downwards

3. From a FBD, you should be able to see that torque due to gravity opposes the torque due to the force F.

100N(0.5 m) - F*(5/3 m) = 0
F = 30 N

Book Problems:

8. Left support F1 = 8570 N, Right support F2 = 6130 N

10. Set the pivot point at a distance x from the left end.
Ma*g*x - Mc*g*(10 m - x) = 0
x = 3.0 meters

11. 3.3 meters from the adult

18. 1.1 m from pivot on side of lighter boy.

22. F1(10.0 m) – (4000 N)(8.0 m) – (3000 N)(4.0 m) – (2000 N)(1.0 m) – (250 kg)(9.80 m/s2)(5.0 m) = 0, so F1 = 5800 N

You can use the sum of forces in the vertical direction to find the other support force F2 = 5600 N

25. If we make A the support force on the left side, and B the support force for the right side, we can assume that the center of gravity is located a distance x from the right side.

Adding torques: (35.1 kg)g(170 cm – x) – (31.6 kg)g*x = 0,
x = 89.5 cm from the feet


27. T = 2500 N, Fay = 2500 N, Fax = 2500 N

You Torque the last piece of Pi... - Torque and Cross Product

I want to remind you all that your midterm is next Monday - exam business as usual with multiple choice and free response.

Also, PLEASE visit the REACH NYC website at http://www.reachnyc.org to register to receive your awards for any AP exams you will take in May. You must print out a copy of Part A, get it signed by your parents, and bring it in to me by next Wednesday. If you do not get these in by the final deadline, the most you can earn for a 3, 4, or 5 on an exam is $300, $400, or $500.


Solutions:

Page 235:
30.
a) Torque = (45 N)(0.84 m)sin(90) = 37.8 N*m
b) Torque = (45 N)(0.84 m)*sin(60) = 32.7 N*m

31.
a) The net torque is the sum of all torques:

-(35 N)(.1 m) - (20 N)(.2 m) + (30 N)(.2 m) = -1.5 N*m

Page 266.
3. 1764 N*m
4. 1.7 meters

Single Harmonica Mocha (SHM) - Solutions

Questions:
2. Since acceleration is proportional to displacement from equilibrium, the acceleration will be zero when the displacement is zero. This happens at the equilibrium position.
7. Frequency will stay the same (look at T for a mass on a spring as an example). Doubling the amplitude will quadruple the maximum potential energy (U = .5*kx^2) which will double the maximum speed since max PE = max KE. Doubling the amplitude also double the maximum displacement, which means the acceleration is also double (a is proportional to -x). The total mechanical energy is quadrupled since there is four times as much initial potential energy.

Problems:
8. Since the cord is elastic, we can treat it like a spring and find a 'spring constant' for it. Rearranging the equation for formula of the period for a mass on a spring, k = 213.2 N/m. We can then calculate that the new frequency with the new mass is 3.77 Hz. Don't forget that frequency is 1/T!

9.
a) Velocity is max at equilibrium position.
0 + .5*k*A^2 = .5*m*v^2 + 0 at equilibrium
k for spring = 177.7 N/m from the period of a mass on a spring for this problem.
solving the first equation, v = 2.82 m/s

b) energy is conserved
0 + .5*k*A^2 = .5*m*v^2 + .5*k*x^2 with x = .10 m
v = 2.108 m/s

c)
total energy of the system is equal to the initial potential energy = .5*k*A^2 = 2 J

d) For x to be a maximum at t = 0, the function must be a cosine function.

x(t) = 0.15*cos(wt) with w = 2*pi/T and T = 1/3 of a second.

15.
The work done on the spring is equal to the potential energy stored in it:
3.0 J = 1/2*k*x^2 with x = 0.15 m

k = 416.7 N/m

The maximum acceleration occurs when the displacement is greatest. This occurs at the beginning when x = .12 m.

a = -k*x/m = -15 m/s^2, so m = 3.33 kg

31.
a)
T = 2*pi*(l/g)^(1/2) and f = 1/T = .613 Hz
b) Using conservation of energy:
0 + mg*(L - L cos(12)) = .5*m*v^2 + 0
v = .532 m/s
c) total mechanical energy is either equal to the max KE or the max PE.
Emech = .5*(.310 kg)*(.532 m/s)^2 = 0.0439 J

32. This problem is very similar to 31. Use conservation of energy, and the fact that the height above the bottom of the swing is L - L cos(theta).
mg(L - L cos(theta)) = .5*m*v^2
v = (2gL(1 - cos(theta))^(1/2)

Lights, Camera, Momentum Review!

Good evening everyone,

I will explain the details tomorrow in class (sort of) so that you know what is going on, but things are going to again be a bit funny Monday for class. As a result, I will NOT be collecting the assignments Monday. The solutions to the review handout are at this link so please check them out before class Monday. Don't worry about the second to last problem - I omitted it from this year's packet.

What I expect you all to do is schedule a time to meet with me to go over this packet of problems. Please email me and let me know when you can meet with me (10 minutes or so) to go over the problems and your answers. Your midterm is STILL next Monday, so we need to make sure you understand anything you are missing at this point.


That said, I hope you enjoy your 'surprise' tomorrow in class.

UPDATE: I hope you enjoyed class today. Here are the solutions to the part II questions.

Don't be so Impulsive - Impulse and 2D Collisions

UPDATED - Book Problem Solutions are below!

Handout Questions:

2.
a) Since the track is frictionless, momentum is conserved in the horizontal direction.

momentum is NOT conserved in the vertical direction since there is an external force acting on the system of the rock and the cart (gravity).
b)
(5 kg)(1.5 m/s * cos 25) + 0 = (5 kg + 10 kg)*vf

vf = 0.453 m/s

3. Since the system is frictionless (even the ramp) we can say that momentum is conserved.

0 = m1*(4 m/s) - m2*(vf)

vf = .667 m/s

b) m1*g*H = .5*m1*v1^2 + .5*m2*vf^2

H = 0.952 m

On the back of the sheet:
1.
a) impulse = area under graph = 8 N*s
b) Impulse J = m*(vf - vo) so vf = J/m + vo = 5.33 m/s
c) vf = 3.33 m/s

4.
a) Momentum is conserved for the x - direction

m*V + 0 = m*V/2 + M*V'

V' = mV/2M

b)

Book Problems:
Questions:
1. Most objects we commonly observe (sliding blocks, cars, WALL-E) are acted upon by friction, meaning that there IS a net external force acting on most objects. As a result, we tend not to observe momentum conservation in the real world the same way we often do in problems.

2. The momentum is transferred into the Earth, though due to its large mass, we don't tend to notice it. Some of the kinetic energy goes into KE of the Earth, but most goes into thermal energy as our muscles slow us down.


Problems
16.
a) Impulse = m*delta_v = (.045 kg)(45 m/s - 0) = 2.025 N*s
b) Impulse = F_av*delta_t so F_av = Impulse / delta_t = 405 N

17. Impulse J = change in momentum. To find the change in momentum here (since it is a 2D problem) we need to look at components. We can set up and to the right to be the positive directions in y and x.

In x-direction:
-mv*sin(45) - mv*sin(45) = -2mv*sin(45)

In y-direction:
mv*cos(45) - mv*cos(45) = 0

Since this is the impulse given to the ball, the impulse given to the wall is directed to the right with magnitude 2mv*sin(45).

38. This is a totally inelastic collision problem, similar to the first one we did in class.

In x:

m1*v1 + 0 = (m1 + m2)*vf cos(theta)

In y:
0 + m2*v2 = (m1 + m2)*vf sin(theta)

Solving the system of equations, theta = 59.6 degrees, vf = 5.48 m/s

39. Another collision problem - momentum is conserved! Drawing a picture of the collision is important to see what is happening here.

In x:
MaVa + 0 = Ma*v'a*cos(30) + Mb*v'b*cos(theta)

In y:
0 + 0 = Ma*v'a*sin(30) - Mb*v'b*sin(theta)

Solving the second equation, v'b = Ma*v1*sin(30)/(Mb*sin(theta))

Substituting and solving, v'b = .808 m/s and theta = 33 degrees.

72.
Momentum conservation gives that 0 = m1v1 - m2v2. If we say that m1 is the mass with higher KE, then 1/2*m1v1^2 = 2*1/2*m2*v2^2.

Solving the first equation and substituting, we can obtain that m1/m2 = 0.5

Where Worlds Collide... - 1D Collisions

HW Solutions:

handout questions:

1. Vf = 1 m/s, delta KE = -60,000 J

2. We can assume momentum is conserved at the moment of the collision, but no more, since gravity acts on the system of the ball and bullet.

final velocity of the bullet and ball is UPWARDS at 33.33 m/s.

Energy is conserved after the collision, with the initial KE = .5*(.150 kg + .030 kg)*(33.33 m/s)^2 and final U = mgh.

Solving, h = 56.69 m.

3. momentum is conserved:

(4 kg)(5 m/s) + (10 kg)(3 m/s) - (3 kg)(4 m/s) = (4 kg + 10 kg + 3 kg)*v_final

v_final = 2.235 m/s to the right.

From the textbook:

Questions:
8. The bat will be in contact with the ball for a longer time since, to hit it from home plate, it must have its velocity turned around. THis will make more sense after tomorrow's lesson.

11. Since the light body has much less gravity force, it must have a smaller mass than the heavy body. Since it has the same KE, this means its speed must be greater than that of the heavy body.

KE = (1/2)mv^2 = (1/2)*mv * v = (1/2)*p*v

Since the KE is the same for both, but the speed of the lighter object is greater, it means its momentum must go down. thus the more massive object will have a greater momentum, even though it is moving at a smaller speed.

12. If an object has no KE, it must have either zero speed or zero mass. In either case, the momentum must also be zero, so the answer is NO.

For the same reason, the opposite is false as well.

Problems:

32.

We know that the total energy is divided into two parts, by conservation of energy:

7500 J = .5*m*v^2 + .5*(1.5m)*v^2

From momentum, we can obtain that v2 = 2/3*v1.

Substituting, we can obtain that:

7500 J = .5*mv1^2 + (2/3)*.5*mv1^2

in other words, 7500 J = 1x + 2/3x where x = the energy of the smaller piece.

Solving, the smaller piece gets 4500 J, and the larger one gets 3000 J

69.

This problem is similar to #11 from yesterday. We can use newton's 2nd/work energy to find the initial speed of each car. We just have to do it twice.

We can find that Car A was going 13.28 m/s and Car B was going 18.78 m/s.

We know that car B was stationary before the crash, meaning it had no momentum.

m_a*v_a + 0 = m_a*v_{f_a} + m_b*v_{f_b}.

We can solve this for v_a = 22.7 m/s.

Since Car A skidded to this speed, we can again use work-energy to find the initial speed of the car:

-mu*m_a*g*delta_x = 1/2*m_a*(22.7 m/s)^2 - .5*m_a*vo^2

Solving for vo = 26.3 m/s. This, converted to miles per hour, is 58.83 MPH. The driver was speeding!

Momentum and Conservation

As promised, here is the solution to the first problem on the sheet. Since I said to the right was positive, any momentum going to the right is positive.

Note the importance of setting the final momentum of the bowling ball as NEGATIVE in part (b) since it is going to the left.



2. Chippy and Skippy:
Since the ice is frictionless, momentum is conserved. Let's set to the right as positive, and assume that Skippy goes to the right.

P_o = P_f
0 + 0 = (2 kg)(1.5 m/s) + (.4 kg)*v_f

Solving, v_f = -7.5 m/s. The negative sign means that Chippy is going to the left.

3. We can again assume momentum is conserved here, as otherwise, the problem could not be solved.
P_o = P_f
m*v + 0 = (m)(-.5v) + (5m)*v_f

v_f = 0.3v

4.
m_bullet*v_o + 0 = (m_bullet + m_block)*v_f

Since the bullet is stuck in the block, they both go at the same velocity v_f = 0.5 m/s.

Solving for the initial speed of the bullet, v_o = 167.2 m/s

HW solutions:
Question 4. If the rich man threw the bag of gold coins in one direction, since momentum is conserved on the frictionless ice, he would slide in the opposite direction, eventually getting to shore.

The loophole to this story is that the gold coins would eventually reach the shore of the lake. Once the rich man also arrived at shore, he could run around the border of the lake to grab his bag of coins. This wouldn't make for as good of a story, however.

Problems:
3. Skip this question until tomorrow - we had to skim past this in class due to time constraints.

4. We can say that momentum is conserved IMMEDIATELY after the collision.

(95 kg)(4.1 m/s) + (85 kg)(5.5 m/s) = (95 kg + 85 kg)*v_final

v_final = 4.76 m/s

5. Momentum is conserved in the horizontal direction.

(12,500 kg)(18.0 m/s) = (12,500 kg + 5750 kg)*v_final

v_final = 12.33 m/s

6. Momentum is conserved.
(9500 kg)*(16 m/s) + 0 = (9500 kg)*6.0 m/s + m*(6.0 m/s)

m = 15,833 kg

8. This problem can be solved in two pieces. Since the block skidded to a stop, we can use either work-energy principle or Newton's 2nd law to figure out the initial speed of the block.

Using either method, we can obtain that v_o = (2*g*mu*delta_x)^(1/2) = 6.823 m/s.

During the collision with the block, momentum is conserved. We now let the initial speed of the block and bullet become the FINAL speed of the collision between the bullet and the block. (v_f = 6.823 m/s)

m_bullet*v_o + 0 = (m_bullet + m_block)*v_f

v_o = (m_bullet + m_block)*v_f/(m_bullet) = 507.2 m/s

11. Momentum is conserved.

(.013 kg)*(230 m/s) + 0 = (2.0 kg)*v_f + (.1023 kg)*(170 m/s)

v_f = 0.39 m/s

Problems 16 and 17 will make more sense later on - don't worry about them for now.

While you're waiting...

Aside from grading exams this weekend, I've been playing around with this GREAT game called Fantastic Contraptions. I recommend you all try it and see what you can come up with. Feel free to email me links of any designs you are especially proud of!

Remember that the lab is due on Monday in class.

This is a site for doing all sorts of fun things (including regressions) on the TI-83:
http://score.kings.k12.ca.us/lessons/ti83tutorial/mainpage.html


Here is the site for regressions from before - the only difference for quadratic regression is that you will have to select 'Polynomial' and order 2. (A quadratic equation is a polynomial equation of degree 2 in x, meaning the 2 is the highest exponent of x.) In short, this will make the regression be a parabola instead of a line.

Oh no!

Hi guys,

I apologize for this, but I left all of the sheets from today at school, and I don't have copies at home to be able to post the answers. To atone, I will be bringing something special for you all for during the test.

Have a good night,

EMW

Work and Energy Crisis - Solutions

A link to the textbook website problems for Chapter 6. The Physlet problems and practice problems are great!

Solutions to today's problems:

1. Using the generalized work energy principle:
Wncf = Wfriction = Kf - Ko + Uf - Uo = .5(2 kg)( 40^2 - 20^2)m^2/s^2 + 0 - (2 kg)(9.8 m/s^2)(80 m) = -368 J

Wfriction = Ffriction*delta_x*cos(180) so Ffriction = -4.6 N

2.
a) a = 40 N / 5 kg = 8 m/s^2
b) Area between F and delta_x = 180 J230 J
c) Wnet = Kf - Ko so Kf = .5*m*vf^2 = Wnet + Ko

So v_f = (2(Wnet + Ko)/m)^(1/2) = (2(180 J + 22.5 J)/5 kg)^(1/2) = 10.04 m/s

d) For it to stop at x = 9 meters, the final KE must be zero. This means that the net work from x = 7 to x = 9 must be equal to Kf - Ko = Ko, the kinetic energy at the speed we found in (c).

The net work is going to be the sum of the work of the net-force and the work of friction, given by Ffric*delta_x*(-1).

The work done by the force from x = 7 to x = 9 is .5*2m*40 N = 40 J.

40 J - Ffric*(2 m)*(-1) = -.5*(5 kg)*(10.04 m/s)^2

Ffric = 106 N

3.
a) With K + U = 25 J, and at a displacement of 4 meters, U = 16 J. Thus K = 9 J

b) K = 1/2*m*v^2 so v = (2*K/m)^(1/2) = 3 m/s

c) When the object has maximum speed, it has maximum KE. This means it must have minimum U, and U has a minimum value of 0 here.

Therefore K_max = 25 J = 1/2*m*v^2

v = 5 m/s

d) The maximum displacement will occur when KE = 0, which occurs when U = 25 J. This occurs at x = +/- 5 meters.

4.
a) W_spring = .5*k*x^2 = 0.2 J
b) Only the spring force does work, so energy is conserved.

Ko + Uo = Kf + Uf
0 + 1/2*k*x^2 = 1/2*m*v^2 + 0
v = 6.32 m/s

c) Wncf = Kf - Ko + Uf - Uo
Ffric*delta_x*(-1) = 1/2*m*v_f^2 - 0 + 0 - 1/2*k*x^2

Ffric = .06 N

The Power of the People

Hope you all enjoyed your election day off. I'm looking forward to some very interesting results tonight as polls close. This has the makings of quite a historic day for all sorts of reasons.

We will start tomorrow by going over the Pre-Lab questions, and then get straight to making measurements with the pull-back cars. Please make sure you go through the steps and think about what we will be doing in class so everything goes smoothly.

We rushed through the last steps to the Mars Rover problem, so here are those solutions, along with the book problem solutions:

Question 7.
e) Since the batteries hold 5.4 x 10⁵ J, and the power required to drive is 10 W, we can figure out how long the batteries will hold up under these conditions:

P = W/delta_t so delta_t = 54,000 s. Since the rover drives at 6.7 x 10^-3 m/s, we can calculate the distance traveled (as we are given the rover land speed) using delta_x = v*delta_t = 361.8 m

f) .010% of the power goes against drag, so this means that .001 W goes against drag. Since the speed is constant, we can use P = F*v to calculate the friction force F = .149 N.

58. P = W/delta_t so delta_t = (285 kg)(9.8 m/s^2)(16.0 m)/1750 W = 25.54 seconds

64. Using Wnet = delta_KE, Wnet = 715.4 J. P = W/delta_t = 357.7 W

67. By drawing a FBD, we can obtain that the force required to move at constant speed must be F = mg*sin(6). The displacement moving up the hill is delta_x, and cos(theta) = 1, since the force is directed along the surface of the hill.

Our equation for power becomes P = F*delta_x/delta_t = mg*sin(6) * v, so v = P / (mg*sin(6)).

The only hiccup is that we need to convert hp (horsepower) to Watts. We can do this using the information on page 169 that says that 1 W = 746 W. This means that P = 186.5 W.

Substituting, v = 2.6 m/s.

71. Since the terminal velocity is 30 m/s, we can assume this is the paratrooper's speed before hitting the snow.
a) Wnet = delta_KE = -36,000 J
b) Wnet = F_average*delta_x*(-1) = -36000 J. delta_x = 1.1 m, so F_average = 32,700 N
c) W_ncf = delta_K + delta_U = .5*(80 kg)*(30 m/s)^2 - 0 + 0 - (80 kg)*(9.8 m/s^2)*(370 m) = -2.54 x 10⁵ J

Generally, I Work Hard by Principle - Energy Problem Solutions

Some solutions to tide you over until class:

From Thursday's handout:

4. Your conservation of energy equation should be:
0 + m₂g*h₂ = .5*(m₁ + m₂ + m₃)*v² + m₂g*h₂ + m₁g*h₂ - m₃g*h₂

This is with h = 4 meters, and v = final speed of all three blocks.

Solving, you should obtain that v = 5.112 m/s.

5. You can set your initial position to be when the block is compressed against the spring, and your final at the max height on the incline. You don't have to find the speed in the middle, though it is ok to do so.

.5*k*x² = m*g*h

h = 16.3 cm

6. You can solve this using a free body diagram, obtaining that T = mg sin(30) = 24.5 N

The other way to do this is using the generalized work-energy principle. The work done by tension here counts as W_ncf since tension is NOT a conservative force.

W_ncf = K_f - K₀ + U_f - U₀
T * x * cos(1) = (0 since constant speed) + mg*x*sin(30)
T = mg sin(30) = 24.5 N

Either method works!

7. It is reasonable to assume (since otherwise there would not be enough information) that the force given is the net force acting on the object.
a) At x = 8 meters, F_x,net = 3 N. This means the acceleration a = Fnet/m = 1 m/s².

b) The work is the area under the graph = .5*(5 m + 15 m)* 3 N = 30 J

c) W_net = Kf - Ko = .5*m*v^2 - 0 so v = (2*Wnet/m)^(1/2) = 4.47 m/s

d) Wncf = Kf - Ko with Kf = 0, and Ko = the answer from b).

Using a FBD, it can be shown that Wncf = mu*mg*delta_x*(-1) = -.5*m*v_f^2.

Thus mu = mv_f^2/(2*g*delta_x) = .204

Problems from Friday:
Worksheet:
Problem 4.
a) v0 = (gR)^(1/2)
b) v = (5gR)^(1/2)
c) Using Wnet = delta_K:
mu*mg*D*(-1) = - .5*m*v^2

mu = 5R/2D

Problem 5.
Wncf = delta_K + delta_U
delta_K = 0 since it is at rest at the start and finish.
delta_U = mg(H2 - H1)

Substituting:
F_fric*delta_x*(-1) = mg(H2 - H1) with H2 = 34 m, and H1 = 38 m.
F_fric = 588 N

Book HW:
Questions:
12. The initial speed of all three balloons is the same. The initial height of all three balloons is the same. The final height of all three balloons is the same.

As a result, since only gravity does work, energy is conserved, and the final speed will be the same in all three situations.

17. At the maximum height, all of the energy is gravitational potential energy. At the moment just before the spring in the pogo stick is compressed, all of the energy is kinetic. At the bottom of the jump, all of the energy is stored in the spring as elastic potential energy. Since gravity and the elastic force are conservative forces, the total energy remains constant.

Problems:
39.
a) v = 9.154 m/s (don't forget to use the initial speed of 5 m/s!)
b) .5*m*v^2 + 0 = .5*k*x^2 + m*g*(-x)
x = 0.362 m

43.
a)
.5*m*v^2 + 0 = .5*k*h^2 + m*g*(-h)
solving, v = 8.03 m/s
b) 3.290 meters

44.
This is the same as the problem we did in class: h = 5/2 R

45.
a) E = K + U = (1/2)mv^2 + 1/2*k*x^2

76.
a) v = (2gL)^2
b) Setting U_g = 0 at the bottom of the swing:
mgL + 0 = .5*m*v_f^2 + mg(2*.2L)
v_f = (1.2*gL)^(1/2)

85.
a) We can use conservation of energy since only gravity is doing work.
.5*mv_o^2 + 0 = 0 + mgL(1-cos(theta))
cos(theta) = 1 - v_o^2/(2*g*L)
theta = 29.3 degrees

b) To solve this, we need to make the observation that Tarzan is moving in circular motion. At the top of the swing, the y-axis must be pointed towards the center of the circle.
T - mg*cos(theta) = 0 (since v = 0, the centripetal acceleration is zero)
T = 640 N

c) Tension will be the greatest at the bottom when Tarzan first grabs the rope. (This is the point when v is greatest, and therefore the centripetal acceleration as well.)

T = mg + mv^2/L = 922.5 N

AP Physics Green - Conserve Energy!

Remember everyone - energy is ONLY conserved if the only work is done by conservative forces. Check this EVERY time you use energy conservation to solve a problem.

Solutions:
36. v_f = 49.5 m/s

37. v_f = (2*g*H)^1/2 = 5.14 m/s

38. (1/2)m^v_o^2 + 0 = (1/2)m^v_f^2 + mgH

v_o = (v_f^2 + 2gH)^(1/2) = 6.45 m/s

40. In each case, you must derive that v_f = (2*g*(H - h2))^1/2

At B: v_f = (2*g*(30 m - 0 m))^1/2 = 24.2 m/s

At C: v_f = (2*g*(30 m - ? m))^1/2 = ANSWER PENDING

At D: v_f = (2*g*(30 m - 12 m))^1/2 = 18.8 m/s

41.
(1/2)m^v_o^2 + mgH = (1/2)m^v_f^2 + 0

v_f = (2gH + v₀^2)^(1/2) = 198.5 m/s.

You do NOT need to use the 45 degrees because v_o is the SPEED of the object at the top!

This has the Potential to be a LOT of work! - Potential Energy

Hello everyone,

This week is going to require that we apply MOST of what we have learned thus far about Newton's Laws and Work/Energy. Working hard is required to really understand how it all comes together.

First, the solutions to the handout problems from class:

1.
a) Determine the magnitude and direction of the force that the spring exerts on the block. [45 N directed to the right.]
b) If the force holding the block against the spring is removed, calculate the work done by the spring on the block. [W_spring = 1/2kx² = 1/2(150 N/m)(0.3m)² = 6.75 J]
c) Calculate the speed of the block when it loses contact with the spring. [Using Work-Energy principle, W_net = KEf - KEi so v_f = 2.598 m/s.
d) If the compression fo the spring is doubled, how would your answer to (c) change? [final speed would double. ]

2.
a) Calculate the displacement of the spring from its equilibrium position. [.049 m]
b) Calculate the work done by gravity as the spring stretches.
c) Calculate the work done by the spring as the spring stretches. [.1801 J]
d) What is the net work done on the mass? [Wnet = 0 since the change in kinetic energy is zero!]
e) How much work was done in lowering the mass to the equilibrium position?

3.
a) Wnet = -24 J
b) Wspring = -1/2kx²
c) Wfric = -mu*mg*deltaX
d) -1/2kx² -mu*mg*deltaX = -24 J so deltaX = .325 m after solving numerically.

Homework Solutions/hints:
Questions:

5. Friction CAN cause an object to accelerate as in the textbook example of ripping out a tablecloth from under dishes. What does this mean about the change in KE? Wnet?

7. Using Uspring = 1/2kx²: (a) Spring 1 - displacement x = F/k. (b) Spring 2

8. KE = 1/2mv², where v is the speed and m is the mass for the object. Neither mass nor v² can be negative, so the answer is no.

9. The amount of work added the same in moving from B to C as compared with the work from A to B. The net work at point C, therefore, is twice the net work at point B. Using the Work-Energy principle, this means that the final speed is the square root of two times vb at point C.

Problems:
11. (a) 1.1*Mg (b) 1.1*Mg*h

14. Using area of a trapezoid: W = .5*(88N/m*.038m+88N/m*.058 m)*(.058m - .038m) = 0.08448 J

29. 0.337 meters

30. mgh = (6 kg)(10 m/s²)(1.2 m) = 72 J

33. a) mgh = (55 kg)(10 m/s²)(3100 m - 1600 m) = 825,000 J
b) minimum work required for the hiker to reach this new height IS the change in PE, or 825,000 J.
c) If the hiker also has a change in KE as compared with the KE at 1600 m, then the work could be more by the work-energy principle.

Net-Work Congestion - An Energy shortage?

7. Any time you are asked to find the minimum work, assume the work is done at equilibrium. Thus the force required to push the cart up the incline in (a) is mg sin(theta). The work is therefore mg*sin(17.5)*300m*(1) = 884,000 J

b) Again, using sigma_Fx = 0, the force required to do this is (mg*sin(17.5) + mu*mg*cos(17.5))* 300 m = 1,600,000 J.

13. The work done by the force is the area under the curve for the given start and end positions.
a) Work = .5*(400 N)*(10 m + 4 m) = 2,800 J
b) Work done from x = 10 to 15 m is NEGATIVE (since it is in the negative direction), so the total work will be 2,800 J - .5*(200 N)*(5 m + 2 m) = 2,100 J

18.
a) If the KE is doubled, the speed is multiplied by the square root of 2.
b) if the speed is doubled, the KE is multiplied by 4.

22. v = (2*Fnet*delta_x/m)^(1/2) = 43.59 m/s

23. Fnet = mv^2/(2*delta_x) = 343 N

25. Through a free body diagram, you should obtain that Fnet = mu*mg. Using this and the work-energy theorem:
mu*mg*delta_x*(-1) = 0 - .5*m*v₀^2
Solving:
v₀ = (2*mu*g*delta_x)^(1/2) = 26.91 m/s

28.
a) This is a Newton's 2nd problem! T = mg + ma = 2479 N
b) Wnet = Fnet*delta_x*(1) = m*a*delta_x = 6791 J
c) W_T = T*delta_x*(1) = 52,059 J
d) W_g = mg*delta_x*(-1) = -45,276 J
e) Wnet = .5*m*v_f² - 0

So v_f = (2*Wnet/m)^(1/2) = 7.852 m/s

Really? You Thought We had no HW?

I wrote this on the board before we started talking about the lab, but yes, we DO have a homework assignment for the weekend!

Chapter 6 - pages 172-179, Question 2 and 3, Problems 1,2,4,5,6, and 8.

Enjoy! Solutions to the HW and quiz will be posted later.

UPDATE:
Quiz solutions are posted here.

Solutions:

1. In this problem, assume the normal force is doing the work in equilibrium, so Fn - mg = 0, so Fn = mg. The work is therefore mg*delta_x* cos(1) = 7350 J
2.
a) Constant speed, so F - Fk = 0. The work done by F = (180 N)(6 m)(1) = 1080 J
b) Again constant speed, so F = mg. W = mg*delta_x*1 = 5400 J

4. The retarding (friction) force does negative work, preventing the car from speeding up. The work done by friction is negative since if the car moves to the right, friction is directed to the left. Thus the only unknown in the equation is the magnitude of the force. F = -70,000J/(2800m*(-1)) = 25 N

5. The only force acting on the rock is gravity. You know how much work is done on it, so the only unknown is the displacement. delta_x = W/mg = 36.1 meters. The signs of the work require a bit more explanation than will make sense here - we can talk more about it in class.

6. The maximum work is what would happen if all of the work done by gravity during the fall went into the nail. This means you have to find the work done by gravity as it falls. This is mg*delta_x*(1) = 7.84 J. This is not very much work! This is why it pays off to exert a bit more force when you are hammering, rather than let gravity do all the work.


7. Any time you are asked to find the minimum work, assume the work is done at equilibrium. Thus the force required to push the cart up the incline in (a) is mg sin(theta). The work is therefore mg*sin(17.5)*300m*(1) = 884,000 J

b) Again, using sigma_Fx = 0, the force required to do this is (mg*sin(17.5) + mu*mg*cos(17.5))* 300 m = 1,600,000 J.

8. Normal force and gravity do no work since they are perpendicular to the horizontal displacement. Since the cart travels at constant speed, the sum of the forces in both directions is zero.

F cos(20) - Fk = 0 so Fk = F cos(20).

The work done by F is F cos(20)* 15 m *(+1) = 169.1 J.

The work done by Fk = F cos(20)* 15 m * cos(180) = -169.1 J.

Around the World in 90 Minutes - Orbit and Planetary Motion

Some distractions before you check your answers:

First, a ride to space on the Space Shuttle:


The view out of a shuttle window during orbit:


This site lists opportunities to see the International Space Station fly over New York City.

There is one pass tomorrow (Thursday) at 6:23 AM, starting in the south at 11 degrees above the horizon, and ending in the East at 35 degrees above the horizon 3 minutes later. It will peak at almost 45 degrees above the horizon halfway through. You will be looking for a fast moving "star", though this "star" is actually a spacecraft housing two Russian and one America astronauts!

Solutions to Orbit Problems from today:

1. The mass of the Earth as determined from this problem is 5.98 x 10²⁴ kg

2. Since the mass of the orbiting satellite divides out from both sides of the equation, it is not possible to determine the mass of a satellite just from knowing its orbital speed and altitude. The government was concerned about what sort of weapon the Russians might be able to load onto such a satellite that passed over the entire world. This kicked off another chain of events that ultimately led to the US developing its own space program.

3. This problem again asks you to derive an expression for the speed of an orbiting satellite. You should obtain that v = (GMe/R)^1/2 where G is the gravitational constant, Me is the mass of the Earth (see question 1) and R is the radius of the orbit of the satellite. The way to increase the altitude of the satellite involves slowing down (or decreasing the speed) by firing a rocket.

In reality, it isn't quite this simple. What is usually done is there are TWO rocket firings required - a first to change the satellite's speed to get it into a new elliptical orbit that reaches the desired new altitude, and a second that pushes the satellite into a circular orbit at the new altitude. This process is called a Hohmann transfer depicted below.


4. Following the steps from class, the radius of the orbit is twice the radius of Earth. The speed of the satellite is 5,591 m/s, and the period is 239 minutes (14340 seconds).

5. minimum speed is (gR)^1/2 = 14 m/s

6. a = (100N*cos(40) - .32*(196 N - 100 N sin(40))/(20 kg) = 1.723 m/s^2

One last toy to check out: An orbit simulator. Try to put some satellites into orbits, and watch how they move differently in elliptical vs. circular paths.

The gravity of the situation - Newton's Law of Gravitation

First, a link to a description of the Cavendish experiment that determined the magnitude of the Gravitational constant to be 6.67 x 10^-11 N*m^2/kg^2.

Homework Solutions:
Questions:
6. The apple exerts a force on the Earth equal to the magnitude of its weight, by Newton's Third Law. This is independent of whether the apple is falling or sitting still. Newton's law of gravitation applies whenever you have two objects with mass, regardless of what they are actually doing.

7. The acceleration due to gravity would be greater since g = GMp/R^2. If the distance was the same (and the orbit was still circular) this would mean that the moon would have a greater speed as it orbits, so it would change phases from new to full more quickly.

16. The question is better answered by saying the gravity force keeps the satellite "down" in its orbit, as it keeps the satellite moving in a circular path. Without the gravity force, satellites would travel in straight line paths.

Problems:
26. using g = GM/r^2, the acceleration is 1.619 m/s^2

30. Calculate the acceleration due to gravity to show the strength of the gravity.

g = 8.938 m/s^2

32. g = .98 m/s^2 = GMp/r^2 so r = 2.017 x 10⁷ m, or at an altitude of 1.379 x 10⁷ m above the surface of the Earth.

34.
a) 9.789 m/s^2
b) 4.346 m/s^2

Curves you can take to the BANK! - Banked Curves

Solutions from the sheet:

1.
a) 22.78 degrees
b) 15.6 m/s

2.
a) T_max = 23.65 N to break
b) T = mg/cos(theta) so theta = arc_cos(mg/T_max) = 78.0 degrees
c) v = (T*L sin(theta)*sin(theta)/m)^1/2 = 8.239 m/s

1995B5
a) acceleration is to the left, velocity is towards the top of the page.
b) .586 m/s
c) v_max = (mu*g*r)^(1/2) = .828 m/s
d) since mass divides out in deriving the result for speed in (c), the presence of the second coin will not change the answer.

Book problems - Chapter 5:
Question 2.
A sharp curve will have a smaller radius of curvature, so the centripetal acceleration will be greater as compared with a curve with a greater radius at the same speed.

Problems:
14.
For the occupants to feel weightless, the apparent weight must be zero. Thus mg = mv^2/R.

You should get that the speed of the riders is 8.57 m/s.

Now you need to convert this to revolutions/minute:

8.57 m/s * (1 revolution/(2*pi*7.5 m))*(60 s/1 min) = 10.91 revolutions per minute.

17.
Calculate the speed required if it is perfectly banked. (12.07 m/s)

Since the 90 km/h (25 m/s) is greater than this, you know the car will slide UP the incline, which means friction must be present and directed down the incline to keep this from happening.

From the x-direction, you should obtain that Fn = (mv^2/R - Fs*cos(theta))/sin(theta).

From the y-direction, you should obtain that Fn*cos(theta) - mg - Fs*sin(theta) = 0.

Substitute the first result into the second, and after some algebra, you can obtain that Fs = mv^2/R * cos(theta) - mg*sin(theta). Substituting the values, you get that Fs = 8035 N.

18.
Your FBD should have Fn towards the center of the circle, mg down, and static friction force up.

In the y-direction, Fnet = 0, so Fs = mg. Since we are looking for the minimum value of mu, Fs = Fsmax = mu*Fn. This means that mg/mu = mv^2/R.

SOlving, mu = gR/v^2 = .198.

There is no force pressing the riders against the wall - the riders feel the increased normal force as a greater apparent weight.

20.
Follow the same steps as for #1 on the worksheet today.
The angle of the incline is 25.75 degrees.
Following the same steps as with part (b) in #1, mu = .765

Running around and falling down - horizontal circles

I will be reminding you of this in class, but in solving these problems, you may NOT just reuse one of the equations we derive in class. On the AP, you will be expected to use basic physics principles to arrive at your answers. You have enough to memorize - do NOT memorize random equations that only apply in specific situations. Remember the physics principles and the relevant equations, and you can derive everything else.

The ONLY new equation thus far in this section has been a = v^2/R for centripetal acceleration. Everything else? DERIVE YOURSELF!

Solutions:
5. The key to this problem is that the hanging block is in equilibrium, which gets you that T = Mg. Your FBD of the disc should get you that T = mv^2/R. Some substitution and algebra gets you the formula given in the text.

6. Draw a FBD first (of course!)

sigma F_x = T = mv^2/R, so v = (TR/m)^(1/2) = 13.96 m/s.

7. Follow the steps from class, as this problem is almost identical. The max speed is 23.43 m/s. In the last step of solving for the maximum speed, the mass divides out, so this result IS independent of the mass of the car!

11. Since the turntable rotates 36 times per second, you can use this to find the speed of the coin. Write the 36 rpm as a fraction: 36 revolutions/1 minute, and then use conversion factors (1 rev = 2*pi*R meters, 60 s = 1 min) to get this to speed v = .4147 m/s.

Since the coin only slides off at this minimum speed, you know that F_s = F_s,max = mu*Fn.

Drawing a FBD, writing Newton's 2nd, and solving gives you mu = v^2/(gR) = .160

16. Your equations for each mass (with T1 as the inner cord, and positive x pointed towards the center of the circle since this is the direction of centripetal acceleration) MUST be:
mass 1: T1 - T2 = m₁v₁^2/R₁
mass 2: T2 = m₂v₂^2/R₂

You can also derive since v₁ = 2*pi*R₁/T (where this T is period, NOT tension), that v₁ = 2*pi*R₁*f. The same thing can be done for v₂.

Thus T2 = 4*m₂*R₂*(pi*f)^2.

and T1 = 4*m₁*R₁*(pi*f)^2 + 4*m₂*R₂*(pi*f)^2

40. acceleration is UP at 3.141 m/s^2.

76. Follow the steps we used for problem 3 in class. v = (g*r*tan(17.5 degrees))^1/2 = 29.15 m/s

Upside-Down antics - Vertical Circles

Remember the advice I gave you in class!

You MUST have one positive axis directed into the center of the circle, so before you start drawing vectors on your FBD, figure out where positive needs to be first.

Centripetal force is the net force in the radial direction of something moving in a circle - use this with Newton's 2nd law to solve the various problems ahead.

Also remember the condition for the minimum speed for something in a vertical circle - the only force acting at this minimum speed is gravity! This is true whrether something is being swung around by a string, riding on a roller coaster, or riding over a hill.

Homework solutions:
6. This is a horizontal circle problem, which we will discuss in class more tomorrow.

8. 3.139 N for the top (a), 9.019 N for the bottom (b)

10. Again, this is a horizontal circle. The key here is that the normal force (the force 'felt' by the trainee' is 7.75*mg, and this is the centripetal force.

Fn = mv^2/R so v = (Fn*R/m)^1/2 = (7.75*mg*R/m)^1/2 = 27.559 m/s

Since one revolution is 2*pi*R meters, this can be converted into revolutions per second using a unit conversion. The correct answer is .439 rev/s.

12. v = (g*R)^1/2 = 9.18 m/s

13.
a) newton's 2nd: mg - Fn = mv^2/R, so Fn = mg - mv^2/R = 5800 N
b) For driver, the equation is the same except that the mass m in the equation is of the driver, not the car. Fn = 406 N.
c) At the minimum speed, Fn = 0. Therefore mg = mv^2/R, and v = (g*R)^1/2 = 31.3 m/s


21. The pilot's path is a circle during the evasive maneuver - in order to NOT hit the ocean during the circular dive, the initial altitude of the pilot MUST be the radius of the path. This is then what the question is asking you to find.

The acceleration must NOT exceed 9g. The radius for this is R = v^2/a_c = 1090 meters.

63. You know that the maximum tension is 1400 N. If you draw a FBD of Tarzan, your equation from Newton's 2nd becomes:

T - mg = mv^2/R

Solving for v, v = (R(T - mg)/m)^1/2 = 6.08 m/s

Going Around in Circles - Uniform Circular Motion

Remember the important parts from today: centripetal acceleration is ALWAYS directed radially (towards the center of the circle) and has magnitude of v^2/R. Uniform circular motion means that the speed v is constant as the object moves around.

Solutions - Problems from Chapter 5:
1. acceleration is 41.667 m/s^2, or 4.252 g's.

2. 1.519 x 10^-2 m

3. speed of Earth around Sun = 2.989 x 10⁴ m/s, so acceleration = 5.954 x 10^-3 m/s^2. We haven't talked about finding the magnitude of the force yet - that will come tomorrow!

Unregrettable Regressions

Hi everyone,

I hope your long weekend is going well - I wanted to post some info that will help you perform a linear regression on your data.

Instructions using excel

Instructions using TI-83/84

Practice before Exam 2!

Here is the link to the website for the book. Pick Chapter 4 and browse around the options at the left side of the screen. The physlet problems and the practice problems are great!

Giancoli 5th Edition Website

Inertial Reference Frames and the Temple of Doom

Questions:
14.
The bag of groceries exerts a 40 N force(a) directed downwards (b) on the person (c).

16. While the forces are the same, the acceleration resulting from a "tug" may be different in both cases. The friction force, which provides an opposing force to the tension in the rope, largely determines whether either side will stay put.

22. By pressing horizontally, the normal force is increased. Since friction force is proportional to the normal force, the friction force directed up the wall will also increase.

Problems:
8. bullet acceleration = 21,875 m/s^2, force on bullet = 153 N.

17. Using g = 9.8 m/s^2, the maximum upwards acceleration is .557 m/s^2. Using g = 10 m/s^2, the maximum acceleration is .357 m/s^2. These should be solved by setting the tension to the maximum value, and then solving the equations of motion for acceleration.

33. You can approximate the two trains connected to a third that pulls it along by having three blocks in a row with two connecting strings:


The masses of the two left blocks are the same. Write Newton's 2nd for both blocks, and you should get that T1 = ma and that T2 - T1 = ma. This should help!

34. This is the same style as the pendulum problem from class. You should be able to obtain from your FBD (remember NOT to tilt your axes!) that T = mg/cos(theta) and that the acceleration is g*tan(theta).

The only catch is that you don't know acceleration. You DO know that it slows to a stop from 20 m/s in 5 seconds. Hint. Hint.

The answer is 21.8 degrees.

58.
a)
This problem works the same as having one block on the table with a string that holds a hanging block (like problem 81!).

For Cleo: T - Fk = ma and Fn1 - m1g = 0
For Figaro: m2g - T = m2a.

Solving algebraically, a = 0.098 m/s^2.
b)
Using kinematics, v0 = 0, a = 0.098 m/s^2, and delta_x = 0.9 meters.
Solving for delta_t = 4.111 seconds.

66. Force required = 21,000 N. Convert 90 km/h into m/s! displacement = 1.042 meters.

79.
Following the logic we used in class, we can derive that a = g tan(theta) = 4.663 m/s^2. From v = v0 + a*delta_t, with v0 = 0, and delta_t = 18 seconds, v = 83.9 m/s.

Complex systems

32.
a) On the FBD of the bucket and girl together, the tension is directed upwards TWICE. The sum of forces in the y-direction becomes 2T - mg = 0 (for constant speed).

Solving, T = 318.5 N
b) The new tension is 350.35 N (1.1* 318.5 N). 2T - mg = ma so a = (700.7 N - 637 N)/65 kg = 0.98 m/s^2.

45.
a) Let F = 730 N and F₁₂ be the force between the two boxes.

Left block:
Fnetx = F - F₁₂ - mu*Fn1 = m₁*a
Fnety = Fn1 - m₁g = 0

Right block:
Fnetx = F₁₂ - mu*Fn2 = m₂*a
Fnety = Fn2 - m₂g = 0


Solving (2), Fn1 = m₁g and Fn2 = m₂g

Substituting:
F - F₁₂ - mu*m₁g = m₁*a
F₁₂ - mu*m₂g = m₂*a

Solving this system, a = 2.476 m/s^2, and F₁₂ = 434 N

61.
a)
We basically derived this in class for the 2nd problem. Let positive be defined to be for m₂ dropping.

a = (m₂g - m₁g*sin(theta))/(m₁ + m₂)

b)
If the system is to accelerate in the positive direction as defined above, then the net force in the numerator of the fraction above must be positive. This will occur if m₂g > m₁g*sin(theta). The system will accelerate to the right if the inequality goes the other way.

69.
This is similar to problem 46 from the other day, as well as to the do-now problem on Thursday. Here, you know mu, delta_x, and are asked to find v0. Check out your work for Problem 46 for guidelines on solving this.

81.
a) The key to this problem is knowing that the block WAS in equilibrium until the last bit of sand is added - from this, you know that Fs = Fsmax = mu*Fn. The hanging bucket (and sand) is in equilibrium, so you can find that T = m₂g.

You should obtain that the mass of the hanging bucket m₂ = 12.6 kg. You must subtract the mass of the bucket to get the sand mass = 11.6 kg.

b)
The equations in the x-direction are now equal to m₁*ax and m₂*ax respectively.

Solving the system of equations, ax = .879 m/s^2.

Problem on page:
acceleration of all three objects is 1.63 m/s^2 in the direction that moves the bucket upwards. The tension is 57.2 N.

The "Inclined Plane" Truth - solutions

46. This problem is the same as our do-now, though it's a little hidden. Stopping distance refers to the distance the block travels as it slows to a stop.
b) 47.37 meters (don't forget to convert 95 km/h into m/s!)
c) the acceleration becomes (1/6) the acceleration due to gravity. If a = 1.63 m/s^2, then the new distance is 284.2 meters.

49.
a) If there is no friction, then a = g sin(theta). If we assume a distance d along the slide (it can be anything!) then we can derive that the speed the child has at the bottom of the slide is V_no-friction=(2gd sin(theta))^(1/2).

If there is friction, the problem says the speed at the bottom is HALF this quantity.
With friction, solve for the new acceleration, and solve for the final velocity of the child at the bottom of the slide. The displacement is the SAME as with no friction.

You should get that a = g sin(theta) - mu*g*cos(theta) and that the final velocity is:
(2gd*sin(theta)-2*mu*gD*cos(theta)) ^ (1/2). If you take this expression and set it equal to 0.5*V_no-friction, you should be able to solve for mu.

The answer (phew!) is that mu = 0.75*tan(28) == .399


51.
a) 1.792 seconds
b) The acceleration does NOT depend on the mass, so the time required would be the same.

52.
a) a = g sin(theta) = 3.67 m/s^2
b) v = (2*a*d)^1/2 = 8.17 m/s

53.
a) 1.226 meters
b) 1.634 seconds (This is just like finding the total time a projectile is in the air - find the time required for the block to stop, and then double it for the entire trip.)

54.

handout answers

1. a = 3.33 m/s^2 DOWN for the 8 kg block.
2.
a) upwards
c) T = 5000 N
d) M = 625 kg

3.
b) T = 1050 N
c) If up is positive, the acceleration of the helicopter is +5.2 m/s^2, and the package is -9.8 m/s^2. (Many of you had this sign wrong when we were working on it on Friday! Your other work should be correct.)
delta_x for the helicopter = 70.4 m
delta_x for the package = 40.4 m
distance = 70.4 m + 5 m - 40.4 m = 35.0 m

4. a = 0.75 g
5.
a) a = F/6m
b) left tension = F/2, right tension = 5F/6

6.
b) F = mu*mg/(cos(theta) - mu*sin(theta))

7.
a) a = 3.0 m/s^2
c) a = 1.96 m/s^2 (NOT the same as part a!)

8. F = g(m1 + m2)/mu

Dynamics solutions set 1

From the handout:

1. acceleration of block = .824 m/s2.
2. Tension in cable = 21000 N, lift force = 72500 N
3. Skydiver acceleration = 5.95 m/s^2 directed downwards
4. Slowing car: acceleration = 1.2 m/s^2, braking force = 1200 N
5. Woman must accelerate DOWN at 2.727 m/s^2
6. a) 980 N, b) 1010 N c) 965 N d) 980 N e) 0 N
7. a) 539 N b) 539 N c) 717 N d) 361 N e) 0 N


Book problems:
Question 11. Although the 2 kg rock has twice the gravity force, it also has twice the mass. By Newton's 2nd law, the acceleration must therefore be the same.

Problems:
12. T = 12960 N

13. accelerates DOWN at 3.5 m/s^2

15. If the thief is able to accelerate down the rope, the tension required to prevent him from falling will be less than his weight, and the "rope" will stay together.

Using the same logic as problem 5 from the handout, the acceleration must be downwards at 2.221 m/s^2 minimum for the rope not to break.

16. accelerates DOWN at .25 g, or at 2.45 m/s^2.

17. Using g = 9.8 m/s^2, the maximum upwards acceleration is .557 m/s^2. Using g = 10 m/s^2, the maximum acceleration is .357 m/s^2.

30.
a) top cord tension = 58.8 N, bottom cord tension is 29.4 N
b) top cord tension = 68.4 N, bottom cord is 34.2 N

39.
a) Since the 40 N force is the minimum force required to get the box moving, the static friction force F_s = F_s,max = mu*F_n. Using a FBD and observing the box is in equilibrium until it starts to move, mu = F/mg = 0.8.

b) F - mu*mg = ma, so mu = (F - ma)/mg = 0.73

44. (After drawing a FBD of course!...) From Newton's 2nd, a = mu*g, and delta_x = -v_o^2/(2*mu*g) = 4.082 meters.

Statics with Friction

From handout:

Problem 1: Fn = 400 N, Fk = 173.2 N, mu = .433
Problem 2: Fn = 250 N, Fk = 173.2 N, mu = .693

Systems Problem (We will talk about this first thing tomorrow. There are some subtle points to discuss on how to do these types of problems.

Mass M = mu*M = 2 kg

Book Problems:
Q22. Since friction force is proportional to the normal force, putting a force against the box and perpendicular to the wall increases the normal force, which then increases the friction force.

Problems:
38. Friction force = 102.9 N. If mu = 0, there wouldn't be any friction force, and the only magnitude of F that could lead to constant velocity across the floor would be zero!

43. 6.667 kg

59. You should obtain that tan(phi) = mu. Thus phi = arc_tan(0.6) = 31.0 degrees.

63.
T = mg = 26.46 N,
Fk = mg - mg*sin(theta) = mu * Fn
Fn = mg cos(theta)

mu = 0.637

Problem 54. The clown must pull with a force of 308.3 N.

Static Equilibrium problems

Some solutions!

Problems:
6. (I will use g = 10m/s^2 to make things easier to calculate.)
a)weight = 200 N, normal force 200 N
b) Table exerts 300 N, bottom box exerts a normal force ON the top box of 100 N.

28.
a)FBD should include F at 45 degrees below the horizontal to0 the right, a friction force directed to the left, and the weight (mg) directed downwards.
b) Friction force F = 88 N cos(45) = 62.2 N
c) Fn = 88 N sin 45 + (14.5 kg)(9.8 m/s^2) = 204.3 N

78.


page 267:
12. T1 (diagonal cord) = 3920 N, T2 = 3395 N
13. right cord = 176.9 N, left cord = 234.8 N
16. tension = 48.2 N. The tension is so great because of the small angle. 2T sin(theta) = mg. Since theta is small, T is large compared with the weight mg.

Review Solutions

Solutions to the MC questions

Pages 1 - 3:
1 E
2 C
3 E
4 C
5 A
6 D
7 B
8 D
9 B
10 D

Page 4:

Page 5 - 6
1. B
2. E

4. E

6. D
7. D
8. C

More Projectile motion!

The written solutions to the handout problems are located at this link.

Robot problem:
Acceleration = -0.5 m/s^2, displacement = 27 meters9.0 meters.

We can look at the other solutions to the handout problems tomorrow in class. I will post others when I can.

Projectile Motion

Question 11. Since the vertical component is zero at the top of the parabolic path, and the horizontal component is the same throughout, the minimum speed occurs at the point of maximum height.

Problems:

20. Cliff is 44.1 meters tall, lands 4.8 meters from the base of the cliff.

22.
The pebbles have a constant horizontal velocity, and zero vertical velocity when they hit the window.

Since the vertical displacement is 8.0 meters, the initial vertical speed is 12.52 m/s. This means (from v = v0 + a*delta_t) that the time to hit the window is 1.277 seconds.

Since delta_x = 9.0 meters, the horizontal component of velocity = 9.0 m / 1.277 seconds = 7.05 m/s. This is the speed of the pebbles hitting the window.

24. The drop takes 3.38 seconds using just the vertical direction. The horizontal speed (which stays constant) is 45 m / 3.38 seconds = 13.31 m/s.

26. It takes 1.228 seconds for the ball to reach its maximum, so it takes double this, or 2.456 seconds to fall back down to the ground.

27. The drop takes 1.621 seconds (since delta_x and vx are given.) Voy = 0 since it is thrown horizontally. The height delta_y of the building is then 12.88 meters.

28. You can set up an equation using the equation for delta_y and time.

-2.2 m = (14 m/s)sin(40)*t - .5*(9.8 m/s^2)*t^2

There are two solutions (times) when this is true, but only the positive root matters: this is at 2.055 seconds. The horizontal displacement is given by (14 m/s)*cos(40)*2.055 seconds = 22.04 meters.

Problem 5 from the handout:
vo = D/(2H/g)^1/2

WP3. Both bullets hit the ground at the same time.

Vector Kinematics

Two important things to keep in mind - keep components separate from each other in a kinematics problem. Second, don't subsitute values until you are ready to get an answer. Keeping things algebraic makes things much cleaner.

Handout Solutions:
5.
a) delta_t = v₀ sin(theta) / g
b) delta_y = (v₀ sin(theta))^2/2g

6. displacement is 65.9 km at [E 15.6 degrees N][E 17.0 degrees N]

7. x-component: 27.19 m/s, y-component: 12.67 m/s

8. 43.01 m/s at 54.46 degrees above the x-axis

9.

10. 60 mi/h 34.28 mi/h

11.
a) 1.72 seconds
b) 14.58 meters
c) double the answer in (a) of 3.44 seconds
d) 124.7 meters

12. Check your notes for these definitions.

Book problems:

Question 10. No - the magnitudes may be the same, but the directions are different. This means the velocities are not the same.

9.
a) West component: 785 km/h* sin(38.5 degrees) = 488.7 km/h (Note that this would be negative according to our usual coordinate system of positive to the right.)
North component: 785 km/h* cos(38.5 degrees) = 614.3 km/h
b)
West: 1466.1 km
North: 1842.9 km

16.
a) Vertical component of acceleration = 3.8 m/s^2 * sin(30) = 1.9 m/s^2
b) delta_y = v0y*delta_t + 1/2*a_y*delta_t^2, but since she starts from rest, v0_y = 0. The delta_y is the elevation change of 335 meters.

Thus delta_t = (2*delta_y/a_y)^1/2 = 18.77 seconds

Vector Mathematics

For those of you finishing the video lab, here is the address for the tracker program:

Tracker:

Video File:
http://groups.google.com/group/bronx-ap-physics/web/legocar.MOV?hl=en

Solutions:

WALL-E Problem Set 1 - Solutions

1. 2.362 x 10⁴ in/min
2.
WALL-E: v0 = 0, delta_x = 0.5*a*(delta_t)^2. Rearranging, delta_t = 11.55 seconds.
EVE: delta_x = v0*delta_t so delta_t = 100 m / (10 m/s) = 10 seconds
3.
WALL-E: Travels 75 meters during the first 10 seconds (from constant acceleration equation used in #2). His average velocity is 75 m / 10 s = 7.5 m/s
EVE: average velocity is the same as initial velocity since her velocity is constant, so average = 10 m/s.
4. Using either kinematics equation and the answer to #2 for WALL-E, the final speed is 17.325 m/s.
5.At time t = 11.55 seconds, EVE has traveled 115.5 meters, and WALL-E has traveled 100 meters. This means that EVE is still 15.5 meters ahead.

WALL-E position function = EVE position function at time t₂ after they have BOTH traveled 100 meters.
100 m + (17.321 m/s)*t₂ = 115.5 m + (10 m/s)*t₂

Solving, t₂ = 2.117 seconds.

Since this occurs after the 11.55 seconds it takes for WALL-E to finish accelerating, WALL-E catches up to EVE after 11.55 + 2.117 = 13.68 seconds.

6.
EVE: 15 seconds.
WALL-E: Must travel 50 meters past the 1st 100 meters. v0 = 17.321 m/s, delta_x = 50 m, so t = 2.887 seconds. Total time of 14.44 seconds.

7. WALL-E will observe EVE moving to the left at 7.321 m/s.

Relating Position, Velocity, and Acceleration Graphs

Check the following link for some solutions to the handout problems.

Problem 7. B and D, the velocity is zero.

Problem 8. Again, at B, D, and E, the object is at rest since the tangent line is horizontal.

Problem 9. Area under the velocity vs. time graph is 640 m.

Problem 10.
a) Moving faster means the speed (or absolute value of the slope) is greater. The slope is greater for object b.
b) Object A has an initial position that is greater than the initial position of object B. As a result, we can define it to be "ahead".
c) The intersection represents when the objects are at the same position, or when they meet.

HW from Textbook: Pages 41 - 46, Questions 5, 18, 19 and Problems 39, 58, 59, and 61
Solutions:
Questions:
5. yes - to win, the car must reach the finish line in the shortest possible time, NOT at the largest possible speed. A car could drive slowly for a while, and then suddenly accelerate at the end to move faster than the other cars.
18. constant positive velocity for 20 seconds, speeding up in positive direction for 15 seconds, slows down and stops at at t = 38 seconds, turns around and speeds up until 45 seconds, and then slows down to a stop at t = 50 seconds.
19. Constant acceleration from t = 0 to 50 seconds, reaches its highest speed at t = 50 seconds, and then slows down and stops from t = 90 to 105 seconds. it then speeds up a bit. Notice that the object NEVER turns around - its velocity is always the same sign.

Problems
39. If we define down to be positive, then the velocity vs. time graph will look like a straight line with positive slope equal to the acceleration of gravity. The position vs. time graph will be a parabola that opens upwards.

58. Check back later for this.

59.
a) negative direction (negative slope)
b) speeding up - the velocity is becoming more negative.
c) negative - the velocity is becoming more negative.
d) positive direction (positive slope)
e) speeding up - slope is becoming more positive.
f) positive - slope is becoming more positive
g) object is not moving from C to D, so its velocity is constant and zero, and therefore acceleration is zero.

61. six times the distance as on Earth

Graphical Kinematics

First, here is a site talking a bit about a VERY exciting event happening Wednesday in Geneva, Switzerland. The largest particle super-collider (considered the most expensive science experiment ever built) will be turned on and tested tomorrow. This is a very big deal - read about it, and also check out this rap below that does a good job of showing what it will be able to do:



Solutions from the handout:
5.
a) 4 m/s²
b) Speeder displacement = 300 meters, police displacement = 200 meters
c) 100 meters
d) 20 seconds (They meet 10 seconds after the police officer stops accelerating, which is 10 seconds after the problem starts.)

6.
b) 4.5 seconds
c) The acceleration is the same at all three times, and has magnitude .267 m/s²
d) Area under the velocity vs. time graph during the acceleration = 1.2 meters
e) The total displacement during the trip is the total area under the curve. Since the area above the t-axis is the same as the area below, the net area is zero.

6(the second one)
a)5 m/s²
b) 1.25 m/s²
c) -2.5 m/s²
d) -5 m/s²
e) 0 m/s²

7.
a) Cart is at rest when velocity is zero, at t = 4 seconds, 18 seconds
b) The cart speeds up when the velocity and acceleration have the same sign. This occurs between t = 4 to 9 seconds, and t = 18 - 20 seconds
c) Displacement from t = 0 to t = 9 seconds is 1.6 meters - 2.5 meters = -0.9 meters, as found from the area above and below the t-axis. Since this is the displacement, and the displacement is x - x0, x = displacement + x0 = 2 meters + -0.9 meters = 1.1 meters.

8. velocity before entering the water is 8.85 m/s, which becomes v0 for the acceleration in the water. Applying v^2 = v0^2 + 2a*delta_x, a = 19.58 m/s².