Problem set #1 solutions.

All from chapter 1.

(You have just started to coast down a hill on your bicycle. The hill has a very steady slope so that you descend 1 m for every 5 m of travel along the hill.)

1. Since energy is conserved, the work needed to be done to pull the object to the top of the hill (if you were to do it) is independent of the path. Straight up, the work is (weight) times (1 distance unit). Up the hill, the work is (unknown force) times (5 distance units). Since they must be equal, the unknown force is 1/5 of the weight. If you’re not pulling the object up the hill, this unknown force is unbalanced, and is what accelerates the object down the hill.
2. Since gravitational acceleration is independent of mass, and the force is 1/5 the weight, the acceleration is 1/5 times g. (g = 22 mph/second or 9.8 m/s/second) which gives 1.96 m/s increase in speed during one second, or 4.4 mph. [OR if you like this better, the force = weight/5 = mg/5, and acceleration = force/m. When combined, this gives acceleration = g/5.]
3. Your velocity increases by another 1.96 m/s or 4.4 mph.
4. During the 2^nd second, your average speed is higher, so you go further.

(A high-jumper and a long-jumper are both human falling objects, but they have slightly different goals.)

1. If the high-jumper jumped directly upwards while not moving horizontally, she would go straight up and straight back down, and would not go over the bar. She needs at least a little horizontal velocity to carry her over the bar.
2. The long-jumper must jump at least partly upward, otherwise he would not spend any time in the air. The time in the air is determined exclusively by the vertical motion. Why not push off at an angle? That would gain him a bit of extra horizontal speed, but at a huge loss to his vertical motion. He would then spend less than the maximum time in the air, and his distance would be less.
3. Since the standing long-jumper has neither horizontal nor vertical velocity to begin with, she must gain some of each. It turns out that dividing them equally, by jumping at a 45 degree angle, is the best.
4. Less gravity means you can both jump higher (good for the high-jumper) and spend more time in the air (good for the long-jumper).

(Three children are at a playground, playing with a seesaw. Two of the children are 3 years old and each has a mass of 15 kg; the third child is 7 years old and has a mass of 30 kg. The seesaw’s pivot is in the middle of the board, and the board itself has essentially no mass. The two 15 kg children sit together at the east end of the seesaw and the 30 kg child sits at the west end.)

1. If the seesaw is balanced, thus experiencing no net torque, it can only move at a constant angular speed. This speed may be zero (if they haven’t started moving yet) or bigger (maybe 20 degrees per second) after someone pushes off but before the opposite end hits the ground.
2. If the 30 kg child pushes on the ground, now there’s an unbalanced torque. The seesaw will change its angular speed (how it will change depends on whether the seesaw was rotating before the push was initiated and in which direction).
3. If one of the 15 kg children jumps off, there is an unbalanced torque. If this happens while the board is both horizontal and stationary, it will rotate faster and faster until the 30 kg end hits the ground. If it happens while the 30 kg end is moving towards the ground, it will speed up. If it happens while the 30 kg end is moving away from the ground, it will slow down, stop, reverse and speed up.
4. If the 15 kg child stays at the east end, then in order to balance the seesaw, the 30 kg child must reduce his lever arm by half. Therefore he should sit halfway between the west end and the pivot.
5. The other 15 kg child returns and sits on the pivot. This has no effect because his lever arm is zero, so he causes no torque.

(Revolving doors—most have four panels arranged in a cross as viewed from above.)

1. It is easier to rotate the door by pushing far from the central pivot because "far from the central pivot" means a larger lever arm. Thus the force you exert is translated into a bigger torque. (Remember for 90° pushing, torque = force times lever arm.)
2. Since you are causing an unbalanced torque, the door undergoes angular acceleration. If it starts at rest, its angular speed increases as time progresses.
3. If there’s no unbalanced torque (which is true after you stop pushing if there’s no friction), the angular speed remains constant. Thus it will keep rotating forever. This is also known as "the law of inertia" or "the law of rotational inertia" or "Newton’s first law (regular or of rotation)."
4. If there is friction, the frictional force itself causes a torque, always opposing the motion. The angular speed then decreases.
5. The mass should be minimized on the outside edge (farthest from the pivot). Here’s why: If there’s a lot of mass on the outside edge, that mass is moving faster than it would if it were near the pivot (think about it—the outside of a wheel moves faster than the inside). Since it’s moving faster, it has more kinetic energy. To give it more kinetic energy, you have to do more work, which is bad.

(Sledding makes use of the nearly frictionless nature of snow. Also assume no air drag. The smooth hill rises 1 meter for every 5 meters you travel along its surface.)

1. If they are travelling at a constant velocity, the total force (same thing as net force) on them must be zero. Since acceleration is zero, the unbalanced force is zero.
2. The uphill force you are exerting (which is only one of the forces involved here—the others are gravity and the "normal" force [the hill pushing]) is 1/5 the weight, or 80 Newtons. The reason is the same as Case 5a above.
3. The (vertical) height of the hill is 50 meters, so the length of the hill is 250 m. Work = force times distance. So work = (80 N) times (250 m) = 20,000 Newton meters, or 20,000 Joules.
4. If the vertical height stays the same, it doesn’t matter how steep the hill is, the work done is the same. For example, even if the sled is pulled straight up, the distance is then 50 m, but the force is 400N. Multiply these together, and you get the same amount of work.
5. Since energy is conserved (remember there is no friction so there is no loss to heat), the work done = potential energy gained. Thus the child/sled gains 20,000 Joules of PE.
6. Again, energy is conserved, but this time it’s converted to kinetic energy. So they have 20,000 Joules of KE when they reach the bottom of the hill.
7. The sled now moves on to flat snow, and the sled slows abruptly. The child falls off the front because there is no force (or not enough force) to slow (decelerate) him at the same rate as the sled. In other words, he has inertia, and so wants to keep going at the same velocity.