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.

Free Fall and other Animals

HANDOUT PROBLEMS:
4. 706.6 meters
5. 4.9 meters
6. 400 meters
7. 1.078 seconds
8. 7.877 m/s

HW SOLUTIONS: Page 43,Problems 34, 37, 38, 41, 44, 46

34. 60.03 meters

37. From the max height, you can get the initial velocity of the kangaroo. (7.275 m/s) You can then use this with a kinematics equation to get the time required to reach this height. Double it to get the answer of 1.485 s.

38. Since the total time is 3.3 seconds, it takes 1.65 seconds to reach the top. You can use this to find the initial velocity of the ball (16.17 m/s) from v = v0 + a*delta_t. The height can then be found using v^2 = vo^2 + 2a*delta_x. The answer is 13.06 meters.

41. Set up a quadratic with delta_y = -105 meters, v_o = 5.5 m/s, and a = -9.8 m/s². Solving numerically, delta_t = 5.224 seconds.

44. You can use v^2 = vo^2 + 2a*delta_x to find the initial velocity　of 12.837 m/s. Again, set up a quadratic to find the times. There are two times because the object has the 12 meter height both when it goes up and goes down. These times are at 0.7309 s and 3.351 seconds.

46. Follow the hint given on the sheet - first try to find the initial velocity of the stone at the top of the window. Since acceleration is constant, we can use the equation stating that average velocity = (v + v0)/2 = delta_x/delta_t. This means the velocity of the stone halfway through its fall past the window is 7.333 m/s. This means that after 0.15 s (halfway of the fall time), the speed of the stone is 7.333 m/s.

For the interval of time from when the stone passes the top of the window to when its speed is 7.333 m/s, v = 7.333, delta_t = 0.15 s, a = 9.8 m/s² downwards, and v0 is unknown. Calculating, v0 = 5.863 m/s.

Now consider the interval of time between when the stone is dropped and it FIRST reaches the top of the window. During this interval, v0 = 0 m/s, a = 9.8 m/s², and v = 5.863 m/s. From these quantities, you can use ONE of the kinematics equations to find the displacement of the stone over this interval. The final answer is 1.754 meters.

Constant Acceleration Equations

Answers from the problems on the back of the handout:

1. delta_x = 300 m
2. final velocity = 8 m/s
3. initial speed = 6.61 m/s
4. t = 0.1714 s
5. time to meet = .636 hrs, position is .184 miles west of flagpole.

HW solutions:

16. acceleration = - 6.25 m/s² (in opposite direction to initial velocity.) This is the same as .638 g's.

21. delta_x = 150 m

23. delta_x = 62.5 m

25. displacement = -156.25 m, time = 25 s

28.


29. time = 23.96 seconds, speed = 66.61 m/s

Velocity and Acceleration as Functions of Time

HW: Read Sections 2-1 to 2-4, Do QUESTIONS 2 - 4 and 10 and PROBLEMS 5, 8, 9, 13

Questions:
2. Velocity is a vector. An object could travel at constant speed but change direction (as in an object traveling in a circle) which means the velocity changes.

3. No - constant velocity means constant magnitude (speed) and direction.

4. No - if velocity is constant, then the average and instantaneous velocities are equal during that time interval.

10. If down is positive, then an object thrown upwards will have a negative initial velocity with a positive acceleration downwards. By switching up to be the positive direction, the reverse will be true.

Problems:
5. Part I of trip takes (130 mi)/(65 mi/h) = 2 hrs. The trip takes 3 1/3 hours total, so you drive at 55 mi/h for 1 1/3 hours. (55 mi/h)(1 1/3 hrs) = 73.3 miles.
a) total distance = 130 mi + 73.3 miles = 203.3 miles
b) average speed = total distance / total time = (203.3 mi) / (3 1/3 hrs) = 60./9 mi/h

8. average speed = 10.37 m/s, average velocity = 3.46 m/s

9. We have to write position functions for each car. Set x = 0 at train A at t = 0. This makes X_a,o = 0 and X_b,o = 8.5 km.

Train A: X_a = X_a,o + v_a*delta-t
Train B: X_b = X_b,o - v_b*delta-t (negative since train B travels in the opposite direction to Train A.

When the trains meet, their position functions are equal.

Solving, delta-t = .0447 hours or about 2 minutes and 41 seconds.

13. 4.256 m/s²

Unit Systems - HW and Solutions

HW for tonight: Read p. 8-14, do PROBLEMS (not questions) 15, 17, 20, 26, 33.

Solutions:
15. You should do this for your own height, but Mr. Weinberg is 5'7", or 67 inches. This works out to 1.702 meters.

17.
a) 3.9 x 10^-9 inches
b) Conversion factor is that 1 atom = 1.0 x 10^-10 meters. You then use the 1 cm length given to find how many atoms there are. Answer: 10⁸ atoms.
20. 7.3% longer

26. There are many correct answers! Here's a reasonable way to go about it.
Say your heart beats 64 beats per minute, and that the average time a person lives is around 75 years. 64 beats/min * (60 min/hr)*(24 hr/ 1 day) * (365 days/year)*(75 years/lifetime) = 2.523 x 10⁹ beats per lifetime.

33.
a) 31,536,000 seconds/year
b) 3.154 x 10¹⁶ nanoseconds
c) 3.171 x 10^-8 years

WP1. units of A: m/s⁴

Welcome back!

Hi everyone,

Glad to know you found the site. I'm very excited to get going this year with you all.

As promised, I'd like you to respond to the following questions for HW tonight:

1. What are you MOST interested in learning during the course?

2. What are you most concerned about in taking this course?

3. What are your strengths and weaknesses as a math and science student?

I ask that you send your responses by email ASAP.

See you all tomorrow!

EMW