Physics 105 - How Things Work - Fall, 1999

Midterm Examination

Given Wednesday, October 13, from 1:00 PM to 1:50 PM

PART I: MULTIPLE CHOICE QUESTIONS

Please mark the correct answer for each question on the bubble sheet. Fill in the dot completely with #2 pencil. Part I is worth 67% of the grade on the midterm examination.

Problem 1:

When you are out fishing on a lake one day, you get your line snagged on a floating tree branch. Luckily the branch is not attached to anything, so you can reel it in to unhook your line. As you reel in the branch, it moves toward you at a constant speed. The amount of work you are doing on the branch is

(A) exactly zero.

(B) constant, but negative—the branch does work on you.

(C) constant and positive—you do work on the branch.

(D) exactly equal to the kinetic energy of the branch.

Answer: (C) constant and positive—you do work on the branch.

Why: You are pulling the branch toward you and it is moving toward you. Although the branch is not accelerating and is thus experiencing a net force of zero, you are pulling it toward you while it is being pulled away from you by drag forces in the water.

Problem 2:

Your pet hamster has an exercise wheel in his cage. He can climb into this wire wheel and then run in it as though he were on a treadmill. When he is running in the wheel, he remains stationary at the bottom of the wheel while it spins around him. In this case his angular velocity is

(A) equal in magnitude to the wheel's angular velocity, but in the opposite direction.

(B) zero.

(C) increasing as he does work on the wheel.

(D) equal in both size and direction to the angular velocity of the wheel.

Why: Since the hamster isn't moving at all, his orientation relative to the wheel's pivot isn't changing. With a constant angular position, the hamster has no angular velocity.

Problem 3:

The county fair has a game that supposedly measures your strength. When you pound on one end of a lever with a giant hammer, the lever tosses a metal ball up a tall, vertical track toward a bell located 10 meters above you (see Figure E). You decide to give it a try and, to your delight, manage to hit the bell on the first try. As the ball rises upward on the frictionless track, it experiences

(A) a constant upward force that depends only on the momentum of the ball as it rises toward the bell.

(B) an upward force that decreases steadily but yet remains substantially greater than zero even when the ball hits the bell.

(C) an upward force that decreases steadily and reaches zero exactly at the moment the ball hits the bell.

(D) no upward forces.

Why: The ball coasts upward because of its inertia alone. There is nothing pushing it upward.

Problem 4:

You are filling a large lightweight dry cleaning bag with helium. At first, the plastic bag doesn't float. But as you keep adding helium to the bag, it eventually begins floating because

(A) at the same pressure and temperature, the upward buoyant force on a helium-filled bag is larger than the buoyant force on an air-filled bag with the same volume.

(B) the average density of the helium-filled bag decreases even though the buoyant force on the bag remains constant as it fills.

(C) as the volume of displaced air increases, the buoyant force increases until it is larger than the weight of the helium-filled bag.

(D) the helium-filled bag's weight decreases as you put more lightweight helium particles inside it.

Answer: (C) as the volume of displaced air increases, the buoyant force increases until it is larger than the weight of the helium-filled bag.

Why: Inflating the bag increases the bag's weight slightly (after all, helium does have a weight, albeit small). But the main effect of the inflation is to increase the bag's volume. As it gets larger, the bag displaces more and more air and experiences a rapidly increasing upward buoyant force. Eventually, the buoyant force exceeds the bag's weight and up it goes.

Problem 5:

You are taking a shower in your dormitory when someone flushes a toilet nearby. The pressure in the cold water line drops and you find yourself showering in what feels like molten lava. This loss of cold water pressure occurs when the flushing toilet lets more cold water flow through the pipes delivering it to the bathroom and the water's speed in those pipes increases. The cold water's faster motion in the delivery pipes reduces its pressure in the shower head because faster moving water

(A) has less pressure than slower moving water.

(B) has less kinetic energy than slower moving water.

(C) has less gravitational potential energy than slower moving water.

(D) losses more energy to viscous drag as it flows through the delivery pipes.

Answer: (D) losses more energy to viscous drag as it flows through the delivery pipes.

Why: If there were no way to lose energy in the pipes, the cold water reaching the shower would always maintain a high pressure and form a strong spray. But as the speed in the cold water pipes increases, the water does lose energy and by the time it reaches the shower head, it just doesn't have much energy left. The hot water then dominates the flow and you get scorched.

Problem 6:

As you ride on a merry-go-round, you feel a strong outward pull that feels just like the force of gravity. This fictitious force occurs because

(A) your velocity is toward the center of the merry-go-round and you experience a fictitious force in the direction opposite your velocity.

(B) you are accelerating away from the center of the merry-go-round and experience a fictitious force in the direction of your acceleration.

(C) you are accelerating toward the center of the merry-go-round and experience a fictitious force in the direction opposite your acceleration.

(D) your velocity is away from the center of the merry-go-round and you experience a fictitious force in the direction of your velocity.

Answer: (C) you are accelerating toward the center of the merry-go-round and experience a fictitious force in the direction opposite your acceleration.

Why: Left to yourself, you would go in a straight line at a steady pace. But the merry-go-round pulls inward on you to bend your path into a circle. As it does this, you accelerate inward and feel your body's inertia trying to make you continue straight. You feel an outward pull, a fictitious force in the direction opposite your inward acceleration.

Problem 7:

Water in a smooth, gentle river flows past the cylindrical post supporting a dock. At the point where this water first encounters the vertical post (see Figure A), the water level

(A) is the same as elsewhere on the river.

(B) rises slightly higher than elsewhere on the river.

(C) drops slightly lower than elsewhere on the river.

(D) vibrates up and down, but averages to the same height as elsewhere on the river.

Answer: (B) rises slightly higher than elsewhere on the river.

Why: As it encounters the leading surface of the post, the water accelerates away from the post. It is beginning its trip around the post. This outward acceleration requires a higher pressure at the surface of the post because this higher pressure is needed to push the water away from the post. This higher pressure also lifts the water level upward and raises the water level at the leading edge of the post.

Problem 8:

You are out in space, so far from any star or planet that gravity is insignificant. You throw two rubber balls so that they drift forward as a pair. These balls continue to touch one another with one ball directly in front of the other. Which of the balls is pushing on the other?

(A) The ball in front is pushing backward on the ball behind and the ball behind is pushing forward on the ball in front.

(B) Only the ball behind is pushing forward on the ball in front.

(C) Only the ball in front is pushing backward on the ball behind.

(D) Neither ball is pushing on the other.

Answer: (D) Neither ball is pushing on the other.

Why: The two balls are coasting forward independently. If either one pushed on the other, they would accelerate in opposite directions and would not continue together. They would drift apart.

Problem 9:

When you blow on a pinwheel, it starts to spin. Even when you aim the air directly toward the wheel's pivot (see Figure C), you produce a torque on the wing-like blades and they undergo angular acceleration. This torque is produced by

(A) lift forces on the pinwheel's blades.

(B) a deflection of the air stream directly away from the pivot.

(C) the buoyant force due to the high pressure air hitting the blades of the pinwheel.

(D) drag forces on the pinwheel's blades.

Why: To make the blades accelerate, something must push them at least partially at right angles to the wind. The blades are evidently deflecting the airstream in order to obtain a sideways lift force and this lift force is what spins the pinwheel.

Problem 10:

You've always wondered how much one of your friends weighs and devise a scheme to measure his weight secretly. You have him sit in a tubular steel chair. This popular style of chair (see Figure F) consists of a single steel tube that's bent into a frame and that supports a seat bottom and a back. The empty chair weighs 10 pounds and is 30 inches tall. The frame acts as a spring and bends downward slightly when the chair is occupied. When you sit properly in the chair yourself, it bends downward 1 inch. When your friend sits properly in the chair, it bends downward 2 inches. From that observation, you know that your friend weighs about

(A) 300 pounds.

(B) twice as much as you do.

(C) four times as much as you do.

(D) 150 pounds.

Answer: (B) twice as much as you do.

Why: The chair is obeying Hooke's law, the concept that a distorted spring experiences a restoring force that is proportional to the extent of its distortion. If your friend bends the chair twice as far as you do, the chair exerts twice as much restoring force on your friend as on you. That means that your friend weighs twice as much as you do.

Problem 11:

When playing hockey, you flick the puck across the ice rink with your stick. The puck then moves across the frictionless surface of the ice at constant velocity. As the puck moves across the ice, its kinetic energy is

(A) constant because kinetic energy is a conserved quantity.

(B) zero because it isn't accelerating.

(C) increasing as the ice does work on it to support it against the force of gravity.

(D) constant because its speed is constant.

Answer: (D) constant because its speed is constant.

Why: Since nothing pushes on the puck, it doesn't accelerate and coasts forward at constant speed. While it thus maintains a constant kinetic energy as well, kinetic energy isn't itself a conserved quantity. Only total energy is conserved. After all, if the puck were to coast uphill on sloping ice, its kinetic energy would decrease even though its total energy (kinetic plus gravitational) would remain constant.

Problem 12:

When a log is floating on water, much of the log is above the water and is actually surrounded by air. If that surrounding air were to suddenly disappear, the log would

(A) float at the same height as before the air left.

(B) move downward slightly and float somewhat lower (deeper) in the water.

(C) sink to the bottom of the water.

(D) move upward slightly and float somewhat higher (less deep) in the water.

Answer: (B) move downward slightly and float somewhat lower (deeper) in the water.

Why: The magnitude of the buoyant force on the log is equal to the weight of everything it displaces: water and air. The presence of air at the beginning adds to the buoyant force on the log. When that air is removed, the water must provide all the buoyant force and so the log slips slightly deeper in the water before being fully supported. (Note that removing the air would cause the water to begin boiling. Nonetheless, the log would slip deeper into this now boiling water before it would be properly supported)

Problem 13:

You are jumping up and down on a trampoline, under the watchful supervision of your gymnastics coach. (We've included the coach because trampolines are dangerous and we don't want to be sued for any virtual injuries that might occur during this exam.) Each time you bounce off the trampoline, you stretch its surface and springs. Just before the surface and springs reach their maximum stretch, your velocity is

(A) downward but your acceleration is upward.

(B) downward and your acceleration is downward.

(C) upward but your acceleration is downward.

(D) upward and your acceleration is upward.

Why: The maximum stretch occurs just before you reach the bottom of the trampoline's travel. (Whenever you are above the trampoline, its surface and springs are barely stretched at all. The only time at which the stretch is significant is when you are pushing the trampoline's surface down hard and are those relatively near the ground.) Just before you reach this lowest point in your travels, you are heading downward (toward that lowest point) but you are losing your downward speed. You are accelerating upward.

Problem 14:

As you stand at the end of a diving board, the board bends downward. While you slowly bend your knees and prepare to jump, the board doesn't move. Finally, when you begin to jump upward, the board

(A) moves upward since the board's restoring force is in the direction of its unbent equilibrium position.

(B) moves downward since you must apply a downward force on the board larger than your weight in order to jump upward.

(C) moves upward since your weight is no longer there to hold the board down.

(D) stays in the same position since the restoring force provided by the spring board balances your weight.

Answer: (B) moves downward since you must apply a downward force on the board larger than your weight in order to jump upward.

Why: To accelerate upward during the jump, something must exert an upward force on you that is larger than your downward weight. The board exerts this large upward force and as it does, it accelerates downward. When it is just supporting your weight, it bends only far enough so that its restoring force equals your weight. During the jump, the forces between you and the board become larger than your weight. The board accelerates downward as you accelerate upward.

Problem 15:

When you pour honey into a bowl, it flows smoothly. If you did the same with water, it would splash. These different behaviors occur because honey's high

(A) viscosity keeps it flowing smoothly while water's low viscosity allows inertia to break its flow into many separate pieces.

(B) momentum keeps it moving in a straight line while water's low momentum allows it to turn abruptly in many different directions.

(C) density keeps it flowing smoothly while water's low density allows it to float upward and splash about.

(D) mass keeps it flowing smoothly while water's low mass allows it to acquire lots of angular momentum.

Answer: (A) viscosity keeps it flowing smoothly while water's low viscosity allows inertia to break its flow into many separate pieces.

Why: Laminar (smooth) flow occurs in fluids when their motion is dominated by the ordering effects of viscosity. The various parts of the fluid push on one another via viscosity and keep the whole fluid moving as an orderly system. Turbulent (swirling) flow occurs when a fluids motion is dominated by the disordering effects of inertia. The various parts of the fluid get ripped apart into a chaotic mess. Honey's high viscosity tends to keep it laminar while water's low viscosity allows it to become turbulent rather easily.

Problem 16:

Your roommate's car has a pair of "fuzzy dice" hanging by a string from its rearview mirror. Yes, they're tacky but you don't have the heart to say anything. Anyway, these dice do a nifty job of indicating how the car is moving. For example, if the dice swing forward toward the car's windshield while the car is on a level road, you know that the car is

(A) accelerating backward.

(B) accelerating forward.

(C) traveling backward at a steady pace.

(D) traveling forward at a steady pace.

Why: When the car accelerates backward, the dice continue at their original velocity for a while and drift forward relative to the car. The car effectively "drives out from under" the dice and they try to leave the car through the front windshield.

Problem 17:

A hot air balloon is open at the bottom and the passengers in the basket beneath can see the inside of the balloon's envelope or "skin." Hot air doesn't flow out of the balloon's open bottom because

(A) viscous forces slow the flow of air out of the balloon.

(B) the balloon's canvas envelope pulls outward, sucking air into the balloon.

(C) the air inside that open bottom is at atmospheric pressure.

(D) the flame pushes the air up into the balloon through that open bottom.

Answer: (C) the air inside that open bottom is at atmospheric pressure.

Why: If the air in the balloon's neck weren't at atmospheric pressure, it would accelerate toward the lower pressure. So for the whole structure to be stable, the neck air is at atmospheric pressure. However, the hot air above the neck doesn't weigh as much as cold air, so its pressure diminishes relatively slowly with increasing altitude. Near the top of the balloon, the pressure inside is substantially higher than the pressure outside the balloon. There is thus an upward force on the balloon and this upward force is the basis of the buoyant force that supports the balloon's weight.

Problem 18:

You are holding two identical-looking balloons, one filled with air and one filled with water. You drop these two balloon from a very tall bridge and notice that the water-filled balloon hits the ground first because its terminal velocity is larger. The terminal velocity of the water balloon is larger than that of air balloon because

(A) conservation of momentum requires the lighter air-filled balloon to travel more slowly.

(B) although the drag forces on the two equally shaped balloons are the same, the buoyant force on the air balloon is larger so the net force on that balloon is smaller and it falls more slowly.

(C) the larger force of gravity on the water balloon must be balanced by a larger drag force, which occurs at a higher speed.

(D) although the force of gravity on the two balloons is the same, the water balloon has more inertia and travels downward more quickly.

Answer: (C) the larger force of gravity on the water balloon must be balanced by a larger drag force, which occurs at a higher speed.

Why: Terminal velocity occurs when the upward drag force on a dropping object exactly balances that object's downward weight. Since drag forces increase with relative speed through the air, the water balloon can obtain the larger drag force it needs by descending faster through the air.

Problem 19:

You are practicing shooting free throws at the basketball court. When you throw the ball, it travels in an arc toward the hoop. Ignoring any forces that air exerts on the ball, the net force on the ball just after it leaves you hand is

(A) down and away from you.

(B) straight down.

(C) zero.

(D) up and away from you.

Why: Only gravity acts on the ball and gravity pulls the ball straight down. Even with air resistance included, the forces on the ball are never forward and thus never point away from you.

Problem 20:

You are swinging a ball on a string around in a horizontal circle above your head and it moves in that circle at a constant speed. We are neglecting air resistance and gravity. Even though the string is exerting a force on the ball, the ball's energy isn't changing because

(A) the negative amount of work that the centripetal force does on the ball is exact cancelled by the positive amount of work that the centrifugal force does on it.

(B) the positive amount of work that the centripetal force does on the ball is exact cancelled by the negative amount of work that the centrifugal force does on it.

(C) the ball's inertia is conserved.

(D) the centripetal force the string exerts on the ball is exactly at right angles to the ball's motion.

Answer: (D) the centripetal force the string exerts on the ball is exactly at right angles to the ball's motion.

Why: The string always pulls the ball directly inward while the ball always moves tangent to the circle of its motion. Those two directions, tangent and inward, are at right angles to one another. Thus the string's force never does work on the ball. It merely redirects the ball's velocity and doesn't change the ball's speed.

Problem 21:

A squirt gun is a simple type of water pump in which a plunger attached to the trigger forces water out of a nozzle and across the room. When you squeeze the trigger of the gun, water squirts out of the nozzle because

(A) the pressure inside the gun is higher than atmospheric pressure.

(B) the water is compressed at the plunger so it must expand out the nozzle.

(C) the Bernoulli effect causes the pressure of water leaving the nozzle to be less than atmospheric pressure.

(D) the Bernoulli effect causes the water's overall energy to increase as it travels through the narrow nozzle.

Answer: (A) the pressure inside the gun is higher than atmospheric pressure.

Why: Squeezing the trigger leads to a pressure rise in the water trapped inside the squirt gun. The high pressure water then accelerates toward lower pressure in front of the gun and sprays out of the gun's nozzle.

Problem 22:

When a modern car crashes into a tree and comes to an abrupt stop, the driver's face and chest collide with an air bag rather than with the steering wheel. The driver's chances of serious injury are reduced by hitting the air bag rather than the steering wheel because the driver transfers

(A) more momentum to the air bag than he would to the steering wheel if there were no air bag.

(B) the same amount of momentum to the air bag as he would to the steering wheel if there were no air bag, but he does so with a larger force because of the air bag.

(C) the same amount of momentum to the air bag as he would to the steering wheel if there were no air bag, but he does so with a smaller force because of the air bag.

(D) less momentum to the air bag than he would to the steering wheel if there were no air bag.

Answer: (C) the same amount of momentum to the air bag as he would to the steering wheel if there were no air bag, but he does so with a smaller force because of the air bag.

Why: The driver is eventually going to stop completely, so the momentum transfer is always the same: the driver gives up all the driver's momentum to the tree. But it's much more "pleasant" to transfer this momentum slowly via small forces than it is to transfer this momentum rapidly via large forces. The air bag slows the transfer and lessens the forces involved.

Problem 23:

You are trying to deliver water to a sink in the tree house in your backyard. You run an old hose from the spigot behind your home, across your yard, and up the tree to the tree house (see Figure B). You let water fill the hose all the way to the tree house sink and then leave the hose pressurized overnight, with no water flowing in it. Unfortunately, the hose can't tolerate high pressure anymore and it springs a leak around midnight. By morning your whole backyard is a swamp. The most likely site for the leak is

(A) on the way up the tree—the most vertical portion of the hose.

(B) on the ground—the lowest point on the hose.

(C) at the spigot—the first point on the hose.

(D) at the sink—the highest point on the hose.

Answer: (B) on the ground—the lowest point on the hose.

Why: With no water flowing in the hose, the water pressure is inversely related to the water's height. The lower the water is in the hose, the greater its pressure (and the smaller its gravitational potential energy). Since the hose is lowest as it passes along the ground on its way to the tree, the pressure in that portion of the hose is greatest. That's why the hose is most likely to burst as it passes along the ground on the way to the tree.

Problem 24:

You're filling in for Wily Coyote in a Roadrunner cartoon. You're trying to tip a huge boulder off a cliff and onto the poor bird. You have just enough strength to start the boulder rocking rhythmically back and forth. To make the boulder rock farther and farther, you should only push it forward when it's

(A) on the far side of its equilibrium position.

(B) rocking away from you.

(C) rocking toward you.

(D) on your side of its equilibrium position.

Answer: (B) rocking away from you.

Why: To add energy to the rocking boulder, you should always do work on it with your pushes. Since you are pushing it away from you, you should only push while it is actually moving away from you. That way, you'll always do work on the boulder with your pushes. If you push on it as it moves toward you, you'll extract energy from the boulder and that will diminish its motion.

Problem 25:

To set the world land speed record and travel faster than the speed of sound, the Thrust SSC vehicle used two jet engines that produced about a 250,000 horsepower. The principal reason why this car needed so much power to travel so fast is that

(A) a car's momentum is proportional to its power, so reaching very high momentum requires very high power.

(B) the pressure drag on a car increases dramatically as the car's speed increases.

(C) the force of the car's momentum was enormous and the jet engines were needed to supply that force.

(D) Newton's second law requires a very fast moving object to have a very large acceleration and thus a very large force.

Answer: (B) the pressure drag on a car increases dramatically as the car's speed increases.

Why: The pressure drag force on a car increases roughly as the square of the car's speed. Since the faster moving car also covers ground more quickly, it does more work against the drag forces each second simply by virtue of covering more distance each second. Taking both the increase in the drag force and the increase in distance traveled per second into account, we find that doubling a car's speed increases the car's power requirements by about a factor of 8. For the Thrust SSC to travel 10 times as fast as an ordinary car, it needs about 103 or about 1000 times as much power as an ordinary car, just to overcome pressure drag forces.

Please give a brief answer in the space provided. Part II is worth 33% of the grade on the midterm examination.

Problem 1:

You are riding your bicycle over a large hill. You have just reached the top and as you start down the other side, you stop pedaling and begin to coast. It's a long, steep hill with a constant downward slope. You are riding a superbly engineered bicycle, which wastes no energy in sliding friction.

(A) As you coast down the hill, your velocity increases. What is the direction of your acceleration?

Why: Your downward weight and the mostly upward support force from the ground combine to create a downhill residual force that causes you to accelerate directly downhill.

(B) At this point, how does the magnitude of your apparent weight compare to that of your actual weight?

Why: Your downhill acceleration is much less than it would be if you simply fell straight downward. The upward apparent weight you would experience if you fell straight downward would be equal in magnitude to your actual weight (which is why you would feel weightless). But the uphill apparent weight you feel as you accelerate slowly downhill is much less than your actual weight. You are simply not accelerating fast enough for that apparent weight to equal your actual weight.

(C) If you continue to coast down the hill, you will eventually stop accelerating and settle in at a constant downhill speed. Even though you are not accelerating, there are still three forces acting on you that perfectly balance each other. What are these three forces?

Why: In the absence of friction, the only three things that push on you are the earth (via gravity), the road (via a support force), and the air (via air resistance or drag).

(D) In what direction do each of these three forces act?

Answer: Your weight (or gravity) acts straight downward, the support force of the road acts at right angles to the road's surface, and air resistance acts directly uphill.

Why: Gravity always acts straight down, support forces always act at right angles to the surfaces exerting them, and air drag acts in the direction opposite the relative motion of an object through the air. Since you are rolling directly downhill on your bicycle through stationary air, the air pushes you directly uphill.

Problem 2:

It's autumn and you are clearing leaves from your yard with a leaf blower. A stream of air rushes out of the blower's nozzle and blows the leaves across the yard.

(A) After the stream of air leaves the nozzle but before it encounters any leaves, what is the pressure of the air stream? (Use words like "more, less, equal, zero" rather than numbers and units.)

Answer: The pressure of the air stream is equal to atmospheric pressure.

Why: If the pressure of the air stream were different from atmospheric pressure, the air around the stream would accelerate it inward or outward. While this sort of acceleration can and does occur at the nozzle, once the stream has left the nozzle, it stops accelerating toward or away from the surrounding air and its pressure is thus atmospheric.

(B) When the stream of air encounters a leaf, what is the pressure of the air stream at the surface of that leaf? (Again, we're looking for comparisons rather than exact values other than zero.)

Answer: The pressure of the air stream is above atmospheric pressure.

Why: When the air collides with the leaf, it accelerates away from the leaf's surface and turns to flow around the leaf. This outward acceleration requires that the pressure of air near the leaf be higher than the pressure far from the leaf. Since the pressure far from the leaf is atmospheric pressure, the pressure near the leaf must be above atmospheric pressure.

(C) You and the blower are transferring forward momentum to the air all the time, yet your momentum remains essentially constant. Why doesn't your momentum change very much?

Answer: You keep transferring any momentum you receive from the air to the ground (via your feet and friction).

Why: The blower does push on your hand because the air is pushing on it. But you don't acquire any net momentum because you get rid of it as quickly as it arrives. You do this by pushing on the ground with your feet. So the air transfers momentum to the blower, which transfers the momentum to you, and you transfer the momentum to the ground.

(D) You're tired of chasing leaves so you begin blowing air across other objects. When you blow a stream of air across the domed lid of a garbage can (see Figure D), that lid suddenly pops upward and then blows away. Evidently, the air stream "sucked" the lid off the can! Use acceleration to explain briefly why the air pressure above the lid dropped below atmospheric pressure. A single, focused sentence is all that's expected here.

Answer: Air accelerates inward as it arcs around the domed lid and this inward acceleration requires that the pressure at the lid's surface be less than the atmospheric pressure far above the lid's surface.

Why: The air is accelerating toward the lid during its travel past the lid. Since that inward acceleration must be caused by a pressure imbalance and is toward lower pressure, the pressure at the lid's surface must be less than the pressure far above the lid's surface. The pressure far above the lid's surface is atmospheric pressure, so the pressure at the lid's surface is less than atmospheric pressure.

Problem 3:

You are part of a team designing an energy-efficient escalator system for a new department store. The store has two floors and patrons will ride between the floors on the escalator. Your team plans to use a single belt of stairs that will travel from the ground floor up to the second floor and then return to the ground floor in a perfectly symmetrical arrangement (see Figure G). The belt will then travel underneath the first floor and reemerge at its starting point. A single motor will turn the belt and convey all of the people up and down between floors.

(A) The belt moves at a very steady pace so that a person riding it upward toward the second floor travels at constant velocity. What is the amount and direction of the net force on that person?

Answer: Zero net force (no direction).

Why: The person is moving at constant velocity and is not accelerating, so the net force on the person is zero.

(B) As that person rides upward toward the second floor, is there any (positive) work being done and, if so, is it being done by the person or by the belt of stairs?

Answer: The belt of stairs is doing positive work on the person.

Why: The stairs are pushing the person upward and the person is moving at least partially upward. Since the force and the direction of motion are somewhat in the same direction (not at right angles and not in opposite directions), the belt of stairs is doing work on the person.

(C) The total weight of patrons on the escalators is 10,000 newtons (about 2,200 pounds). Half the people (weight 5,000 newtons) are riding the upward escalator and half (weight 5,000 newtons) are riding the downward escalator. The belt advances 1 meter each second. Neglecting friction and air resistance, how much power must the motor provide to the belt?