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