• Due Monday, February 2, 1998, in class
• Please answer each problem as concisely as possible
• You may discuss the problems with one another but you must write them up separately and in your own words.
• There are 5 problems containing a total of 24 parts. Each part will be worth 4 points.
• You will receive an additional 4 points for writing your name legibly on the problem set before turning it in.

Problem 1: Chapter 1, Case 1a-e (Pg. 72)

You’re riding on a playground swing. You’re traveling back and forth once every few seconds.

a. At what point(s) in your motion is your velocity zero?

Answer: At the end or peak of each swing.
Why: As the you approaches the end of a swing, you are accelerating directly backward and you slow to a complete stop. You continue to accelerate in that same direction and thus reverse your direction of travel.

Answer: At the end or peak of each swing.
Why: At the end of a swing, you are as high up as you ever get. Since gravitational potential energy is proportional to altitude, this highest point is also the point at which your gravitational potential energy is at its maximum.

Answer: At the bottom or middle of each swing (when you a directly below the support for the ropes).
Why: When you are at the bottom of a swing, you are as low down as you ever get. Since your gravitational potential energy is then at its minimum, your kinetic energy must be at its maximum. Equivalently, your speed is greatest at the bottom of the swing because you have been accelerating partially forward up to that point. Since kinetic energy is proportional to speed squared, your kinetic energy is also maximum. Once you pass the bottom of the swing, you begin to accelerate partially backward and lose both speed and kinetic energy.

Answer: Yes.
Why: At the bottom of the swing, your speed is not changing but your direction of travel is. You are accelerating upward. To see that this is the case, note that your direction of travel is tipping more and more upward throughout your swing, meaning that the direction of your velocity goes from significantly downward near the start of the swing to significantly upward near the end of the swing. For your velocity to tip upward in this manner, you must be accelerating partially upward. At the bottom of the swing, you are accelerating directly upward!

Answer: More than your weight.
Why: Since you are accelerating upward at the bottom of the swing, the net force on you must be upward. This can only occur if the ropes are pulling upward on you with a force that is larger in magnitude than your weight. That way, you are pulled upward harder than you are pulled downward and you accelerate upward. That's how your velocity tips more and more upward so that you follow an upward circular arc rather than falling or heading in a straight line.

Problem 2: Chapter 1, Case 17a-e (Pg. 75)

You’re seated in a crowded bus that has stopped at a bus stop. Two people board the bus, one wearing rubber-soled shoes and one wearing inline skates (roller skates — i.e., no friction). They both have to stand in the middle of the aisle and neither one holds onto anything.

a. The bus driver is new and the bus lurches forward from the bus stop. Which way does the person wearing inline skates move in relationship to the bus?

Answer: The skater moves backward relative to the bus (but stays in one place relative to the bus stop).
Why: As the bus accelerates away from the bus stop, the ground pushes it forward. To remain with the bus, the skater also needs a forward force but the skates prevent the
bus from exerting such a forward force on the skater. The wheels of the skates just spin. So instead of accelerating forward, the skater remains in place in front of the bus
stop and the bus drives out from under the skater. To the passengers on the bus, the skater rolls backward through the isle and hits the back of the bus.

Answer: The bus can't exert the forward force on the skater that the skater needs to accelerate forward with the bus.
Why: The skate wheels turn as the bus accelerates forward, so that the skater doesn't experience the forward frictional force that a person in normal shoes would need to
accelerate with the bus.

Answer: (Static) friction.
Why: Static friction between the bus floor and the person's rubber-soled shoes pushes the person forward and the bus floor backward. As a result, the person accelerates with
the bus and the bus accelerates somewhat more slowly than it would have had the person not been in the bus.

Answer: Zero.
Why: Since the bus and the skater are both moving at constant velocity, neither one is accelerating. The skater (like the bus) is experiencing zero net force.

Answer: The force of static friction can't provide enough backward force to slow the person with the bus.
Why: When the bus stops gradually, static friction between the shoes and the bus floor exert enough backward force on the person wearing the shoes to make that person
accelerate backward with the bus. However, when the bus stops suddenly, the force required to accelerate the person backward with the bus is greater than the maximum
force that static friction can supply. The shoes begin to slide across the floor and the person skids forward in the bus.

You’re about to go on a bicycle trip through the mountains. Being ambitious, you decide to take two children along. The children sit in a trailer that you pull with your bicycle.

a. As you wait to begin your trip, you and your bicycle are motionless. What is the net force on your body?

Answer: Zero.
Why: You are not accelerating so the net force on you must be zero!

Answer: Zero.
Why: You are still not accelerating (your velocity is constant) so the net force on you must be zero!

Answer: 40 N
Why: Since the children and trailer weigh 400 N, lifting them directly upward a distance of 1 m will require 400 N-m of work (400 N of upward force times 1 m of distance in the direction of that upward force). Because energy is conserved, raising the children and trailer 1 m upward through any other path will also require 400 N-m of work (assuming that you increase only their gravitational potential energy, which is why this problem mentions that you do everything at constant velocity). If pulling the children and trailer up a ramp that's 10 m long will raise them 1 m, then you will be doing 400 N-m of work over a distance of 10 m. That requires the force you exert on the children to be 40 N, so that the work you do (40 N of uphill force times 10 m of distance in the direction of that uphill force) gives you 400 N-m again.

Answer: 200,000 N-m (or 200,000 J) and No (it doesn't depend on the path)
Why: The children and trailer weigh 400 N, so raising them upward 500 m will require 200,000 N-m of work (400 N of upward force times 500 m of distance in the direction of that upward force). Any  path to the top of this hill will require you to do the same amount of work on the children and trailer because the work you do simply increases the children and trailers' gravitational potential energy, an increase that doesn't depend on the path taken in raising them upward 500 m.

Answer: -200,000 N-m (or -200,000 J)
Why: To lower the children and trailer down the hill at constant velocity, you must push them uphill while they move down hill. Since the force you exert on the children and trailer is in the direction opposite the distance they travel, you do negative work on them (or, equivalently, they do positive work on you). No matter how steep the hill, the work you do in lowering them a total of 500 m at constant velocity is going to be the same as if you had lowered them directly downward. In that case, the upward force you exert on them is 400 N and the downward distance they move is 500 m. The total work you do on them is -200,000 N-m, where the minus sign comes from the fact that the force and distance are in opposite directions.

Problem 4: Chapter 1, Case 25a-e (Pg. 76)

Advanced skiers turn by sliding the backs of their skis across the snow. Since the fronts of their skis don’t move much, the skis end up pointed in a new direction.

a. The amount of sideways force that a skier must exert on the skis to slide them sideways is proportional to how hard the skis press down on the snow beneath them. Why?

Answer: The skier experiences sliding friction when sliding the skis sideways and sliding friction is always roughly proportional to the forces pressing the two surfaces together.
Why: It is the nature of  friction that the harder the two surfaces are pressed together, the harder it is to slide them across one another. The force of sliding increases almost exactly in proportion to the forces pushing the surfaces together.

Answer: (If the skier is willing to accelerate downward,) he or she can simply not push downward on the snow so hard.
Why: There is no law that requires the skier to push downward on the snow with a force equal to his or her weight. In fact, if the skier is accelerating up or down, they will not be exerting a force equal to their weight on the snow. Acceleration requires a non-zero net force, so the snow must push on the skier with a force that doesn't balance the skier's weight. If the snow pushes upward extra hard on the skier, the skier will accelerate upward. If the snow pushes upward weakly or not at all, the skier will accelerate downward. The skier can choose the amount of upward force the snow exerts on him or her by choosing how hard to push down on the snow. The snow merely responds with an equal but oppositely directed force--in accordance with Newton's third law.

Answer: At the top of the jump (or, more accurately, during the moment of maximum downward acceleration, or, equivalently, during the moment of minimum upward force from the snow on the skier)
Why: At the top of the jump, the skier is accelerating downward rapidly. For this downward acceleration to occur, the snow must not be pushing upward on the skier very hard, if at all. With little or no force pressing the skier's skis against the snow, the skis turn easily because there is almost no sliding friction between the skis and snow.

Answer: Pulling the legs upward separates the skis from the snow so that they can't push against one another very hard (or, more accurately, pulling upward on the skis causes the skis to accelerate upward so that they move away from the snow and don't press on the snow hard).
Why: Lifting the legs suddenly is an effective way of reducing the forces between the skis and snow--if they aren't touching, they can't exert support forces on one another.

Answer: Without unweighting, the force of sliding friction is large. Since the work done against sliding friction is the product of the force needed to slide the skis times the distance of the slide, having a large sliding friction force to overcome makes the work required to slide the skis sideways large.
Why: The more force pressing the skis against the snow, the larger the force of sliding friction that opposes the slide. Since the skier must overcome that sliding friction force in doing the work required to turn the skis, increasing the sliding friction force increases the work done.

Problem 5: Chapter 11, Case 4a-d (Pg. 408)

The spark lighters used in gas stoves and grills are based on small piezoelectric crystals. When you compress a piezoelectric crystal, it produces a charge separation. One side becomes positively charged while the other becomes negatively charged. In a typical spark lighter, a spring-loaded mass strikes the crystal and produces a huge charge separation. Wires attached to the crystal allow these charges to approach one another across an air gap and create a spark.

a. Separating charge takes energy. Show that the spring-loaded mass does work on the crystal while compressing it.

Answer: The mass pushes the crystal's surface inward and the crystal's surface moves inward.
Why: Compressing something always requires work because you must exert an inward force on the object's surface and that object's surface must move inward. Since the force and distance moved are in the same direction (and not zero), you are doing work--force times distance in the direction of that force.

Answer: For charge to jump from one wire to another, it must experience a large force. That force increases as the amount of charge on each side of the crystal increase.
Why: Charges don't jump into the air for no reason--something must push them. For a positive charge to leave a metal surface, that metal surface should have many other positive charges on it and, ideally, there should be another nearby surface covered with negative charges. That way, the positive charge that's going to jump is pushed away from the metal by its fellow positive charges and is pulled out of the metal by the nearby negative charges on the other surface. The more positive charges on the metal surface, the more force the jumping charge experiences and the more likely it is to leave the metal.

Answer: A sharp point packs like charges very close together so that they repulsive forces become very large. These repulsive forces make it easier for a charge to enter the air.
Why: On a smooth, lightly curved metal surface, like charges spread out evenly and experience relatively small repulsive forces. With little pushing them away from the metal, they are unlikely to enter the air. But at a region of severe curvature, particularly at a sharp point, the charges bunch up into a small area and repel one another strongly. They push one another into the air easily.

Answer: As the wires separate more and more, the attractive forces that the opposite charges on the two wires exert on one another decrease and it becomes harder for charges to leave the wires and jump through the air as a spark.
Why: For a positive charge to leave a surface, it needs an outward force. Other positive charges on that surface provide some of this outward force, but the presence of a nearby negatively charged surface helps pull the positive charge, too. As that negatively charged surface moves away into the distance, its attractive forces decrease and the positive charge is less and less likely to jump.