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Instructor: Dr. Scott A. Yost Office: 307 Nielsen Physics Building Hours: Monday, Wednesday and Friday, 1-2 PM or by appointment Textbook: L.A. Bloomfield, How Things Work, 2nd ed. |
Phone: 974-7852 E-Mail: syost@utk.edu Web Site: http://homework.phys.utk.edu/phys101 |
These are the answers to the even-numbered exercises and problems, and all of the cases assigned in Chapter 1. The answers to the others appear in the back of the textbook.
Exercises
2. No force keeps you moving. Inertia does. Inertia is not a force.6. The driver has inertia, and will keep moving in a straight line at his original velocity, until he encounters the steering wheel, which has stopped.
8. Although your speed is constant, your direction of motion constantly changes, which means that the velocity, a vector, is constantly changing.
14. The ball's speed increases the same amount every second: 9.8 m/s.
16. Its average speed is only 4.9 m/s during the fall, so it has fallen only half way to the ground after one second.
18. Kicking the ball straight up would give it the most time in the air.
20. The net force is zero, since she is not accelerating.
30. You do positive work on the kite when the force you exert on it is in the direction of motion. This is when you pull the kite in.
34. You are doing work on the saw in both directions.
38. Gravity slows you down, as your kinetic energy is converted to gravitational potential energy while climbing the hill.
Problems
2. t = v/a = (24.6 m/s) / (4 m/s2) = 6.15 s
6. Use v = at and h = at2/2 = v2/(2g) to find that the squared initial velocity v2 = 2gh = 9.8 m2/s2. Then v = 3.13 m/s .
10. 60 kg x 9.8 m/s2 = 588 N.
14. F = mg/20 = 20,000 kg x 9.8 m/s2 / 20 = 9,800 N.
16. Work = mgh = 20,000 kg x 9.8 m/s2 x 50 m = 9.8 x 106 J
18. PE = mgh = 1000 kg x 9.8 m/s2 x 200 m = 1.96 x 106 J
20. Work = Fd = 2000 N x 0.2 m = 400 J
22. Work = Fd = 30 N x 1000 m = 30,000 J
Cases
1a. At the highest points (turning points) of the swinging motion. 1b. At the highest points (turning points) of the swinging motion. 1c. At the bottom (center) of the swinging motion. 1d. You are accelerating upward, but not horizontally. 1e. The upward force on you is more than your weight, so that you can accelerate upward.
2a. 1.4 s 2b. 14 m/s 2c. More than in part (b). 2d. Your friend will hit the water first. 2e. You will hit the water at the same time. 2f. The person running will hit the water with the most speed..
4a. They will reach the bottom at the same time. 4b. Your toboggan will experience the greater change in velocity, because its mass is smaller. 4c. You would descend faster on a steeper route, because a larger component of your weight would be directed down the hill to provide the acceleration. 4d. The PE decreases going down the hill, and increases again when carried up the hill. 4e. When you pull the toboggan up the hill. 4f. When gravity pulls the toboggan down the hill.
7a. This force is equal to the force of the starting block on the runner which provides the initial acceleration for the race. 7b. The large mass of the shot gives it a lot of inertia, so it requires a lot of force to change its velocity. 7c. Your upward acceleration would be much greater without the support pad, since your velocity would decrease to 0 much faster. The force is proportional to the acceleration, so the force on you would be much greater without the pad. 7d. The net force is backwards, opposing your motion. 7e. A larger upward component of velocity keeps you in the air longer so you can go farther. A 45 degree take-off angle is optimal for distance.
| Department of Physics | University of Tennessee |