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37 Cards in this Set
- Front
- Back
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Cite an example in which a force is exerted on an object without doing work on the object.
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When a person pushes on a wall, the wall doesn’t move, therefore no work is done
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Which requires more work: lifting a 50-kg sack vertically 2 meters or lifting a 25-kg sack vertically 4 meters?
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They both require the same amount of work
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If both sacks in the preceding question are lifted their respective distances in the same time, how does the power required for each compare? How about for the case where the lighter sack is moved its distance in half the time?
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Power required to lift 50 kg sack= 1000N-m/ t - Power require to life 25 kg sack= 1000N-m/t
The power is equal but if the time taken to life the 25 kg sack is t/2 than the power is 2000N-m/t; the 25 kg sack would weigh more |
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A car is lifted a certain distance in a service station and therefore has potential energy relative to the floor. If it were lifted twice as high, how much potential energy would it have?
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Twice
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Two cars are lifted to the same elevation in a service station. If one car is twice as massive as the other, how do their potential energies compare?
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The twice-as-massive car has twice the PE.
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A moving car has kinetic energy. If it speeds up until it is going four times as fast, how much kinetic energy does it have in comparison?
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A three-times-as-fast car has 42 or 16 times the KE.
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Compared to some original speed, how much work must the brakes of a car supply to stop a four-times-as-fast car? How will the stopping distance compare?
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A four-times-as-fast car has 16 times as much KE and will require 16 times as much work to stop, and 16 times as much stopping distance.
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What will be the kinetic energy of a pile driver ram when it undergoes a 10-kJ decrease in potential energy?
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If none of this PE is converted to heat, the KE will be 10 kJ.
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An apple hanging from a limb has potential energy because of its height. If it falls, what becomes of this energy just before it hits the ground? When it hits the ground?
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PE turns to KE in falling; when it hits the ground it goes into deformation and heating the apple and ground.
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What does it mean to say momentum is a vector quantity and energy is a scalar quantity?
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Momentum has direction, is a vector, and can be cancelled. Energy is a scalar, is without direction, and can’t be cancelled.
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If a moving object doubles its speed, how much more momentum does it have? How much more energy?
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Twice as much momentum; four times as much kinetic energy.
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Which requires more work to stop -- a light truck or a heavy truck moving with the same momentum?
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- 1000 kg. x 20 m/sec, or 2000 kg. x 10 m/sec. The momentum - product of mass times velocity is the same in both cases.
-The kinetic energy is (1/2) (mass) (velocity squared); in the first case, that would be 200,000 J, in the second case, 100,000 -So, the light truck has more energy (and requires more work to stop), because it moves faster - and because energy is proportional to the square of the speed. |
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At what point in its motion is the KE of a pendulum bob a maximum? At what point is its PE a maximum? When its KE is half its maximum value, how much PE does it have?
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The KE of a pendulum is maximum at the bottom of its swing where its PE is minimum (or where we often take its PE to be zero). The PE of a pendulum is maximum at the top of its swing where it momentarily stops and its KE is zero. The total energy remains constant, E = KE + PE; So when the pendulum's KE is half its maximum value, its PE is half its maximum value if we have counted its PE as zero at the bottom of its swing.
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A Physics instructor demonstrates energy conservation by releasing a heavy pendulum bob, as shown in the sketch, allowing it to swing to and fro. What would happen if, in his exuberance, he gave the bob a slight shove as it left his nose? Why?
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If the bob starts off from rest, it will convert its PE into KE going down to the bottom of the swing and then convert KE back to PE and come to rest at the other end of its swing and then start back. On the way back, it increases its KE by reducing its PE until it comes to the bottom of the swing. On the last part of its swing, this KE decreases as it rises and increases its PE until the KE finally goes to zero and the pendulum comes to rest. With KE = 0, the PE must be equal to the total energy which is the original PE and it is back at its original height.
However, if the bob starts off with some KE because of a shove, then its total energy is greater than for the case just described. At the bottom of its swing, the pendulum will have a greater speed due to this increased energy. At the other end of the swing, it will come to rest at a greater height so its PE is greater. On the way back, that great height means it will smash our exuberant physicist in the nose! |
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Why does the force of gravity do no work on ? (a) bowling ball rolling along a bowling alley, and (b) a satellite in circular orbit about the Earth.
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For gravity to do work, the motion must be in the direction of the force of gravity. In your two examples, the motion is perpendicular to the force of gravity, so gravity can do no work on those objects.
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Why does the force of gravity do work on a car that rolls down a hill but no work when it rolls along a level part of the road?
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On a hill, the gravity force can be broken down into a force perpendicular to the slope and a force parallel to the slope. The component of gravity parallel to the slope is what causes a car to roll downhill. On level ground, there is only a downward force perpendicular to the ground, hence no motion caused by gravity.
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Does the string that supports a pendulum do work on the bob as it swings to and fro? What about the gravitational force ... does it do any work on the bob?
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The tension in the string is always perpendicular to the direction of motion of the bob and so it does no work. The gravitational force is always directed vertically down and so when the bob it at the mid-point of its swing, i.e., at the bottom, the force is perpendicular to the motion. However, at positions away from the mid-point there will be a component of the gravitational force acting in the direction of motion, which increases to a maximum when the bob is at the top of its swing. Hence, the gravitational force does do work on the bob at all points in the swing except at the bottom.
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. Suppose that you and two classmates are discussing the design of a roller coaster. One classmate says that each summit must be lower than the previous one. Your other classmate says this is nonsense, for as long as the first one is the highest, it doesn't matter what height the others are. What do you say?
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As long as the first summit is the highest, it does not matter what height they are because then the car will always have some kinetic energy to more forward.
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Consider the identical balls released from rest on tracks A and B, as shown. When they reach the right ends of their tracks, which will have the greater speed? Why is this question easier to answer than the similar one (ex. 40) in Chapter 3?
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Both balls begin their journey with the same energy, and both end with the same potential energy (because they end at the same altitude) so they must also both end up with the same kinetic energy (and thus the same speed.)
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If a golf ball and a ping-pong ball move with the same kinetic energy, can you say which has the greater speed? Explain in terms of the definition of KE. Similarly, in a gaseous mixture of massive molecules and light molecules with the same average KE, can you say which have the greater speed?
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Kinetic energy = 1/2 m v2.
Since the golf ball has a higher mass than the ping pong ball, the only way they could both have the same kinetic energy is if the gold ball has a slower speed. For exactly the same reasons, the light gas molecules must be moving faster than the heavy ones, if we want them all to have the same KE. |
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You're on a rooftop and you throw one ball downward to the ground below and another upward. The second ball, afer rising, falls and also strikes the ground below. If air resistance can be neglected and if you downward and upward ititial speeds are the same, how do the speeds of the balls compare upon striking the ground?
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The speeds of the balls upon striking the ground will be the samefor both.
Since conservation of energy gives Since the initial KE and PE are the same for both, the final KE and PE will also be the same. |
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Does the KE of a car change more when it goes from 10 to 20 km/h or when it goes from 20 to 30 km/h?
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The change in kinetic energy, when it goes from 10 to 20 km/h, is =1/2 m (20^2-10^2)= 150m unitts; The change in kinetic energy, when it goes from 20 to 20 km/h is = 1/2m(30^2-20^2); 1/2m(900-400) = 250 units; the kinetic energy is greater in the 2nd case
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Which, if either, has greater momentum: a 1-kg ball moving at 2m/s or a 2-kg ball moving at 1m/s? Which has greater kinetic energy?
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KE of 1kg ball moving at 2m/s = ½ X 1 X 2^2kg-m^2/s^2 = 2J; KE of 2 kg ball moving at 1 m/s = ½ X 2 X 1^2kg-m^2/s^2 = 1J; Therefore, the KE in the first case is greater
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Two lumps of clay with equal and opposite momenta have a head on collision? Is momentum conserved. Is kinetic energy conserved? Why are your answers the same or different?
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Yes, momentum will be conserved. It was zero before collision (as they where equal and opposite so, they cancel out each other) and remains zero after collision. (Momentum is a vector quantity). No, KE will not be conserved but total energy will be. The KE will get converted into heat energy(internal energy of the system will rise at expense of KE)
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Consider the swinging-balls apparatus, also known as Newton's pendulum. If two balls are lifted and released, momentum is conserved as two balls pop out the other side with the same speed as the released balls at impact. But momentum would also be conserved if one ball popped out at twice the speed. Can you explain why this never happens?
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Indeed, momentum would be conserved if a single ball popped out with twice the original, incoming speed. We might write that as
initial momentum = (2 m) (vo) = (m) (2 vo) = final momentum But what does that do to the KE? Remember that KE = (1/2) m v2 initial KE = 2 [ (1/2) m vo2 ] =?= 1 [ (1/2) m (2 vo)2 ] = final KE initial KE = 2 [ (1/2) m vo2 ] =?= 1 [ (1/2) m (4 vo2 )] = final KE initial KE = 2 [ (1/2) m vo2 ] =?= 4 [ (1/2) m vo2 ] = final KE initial KE = 2 [ (1/2) m vo2 ] =/= 4 [ (1/2) m vo2 ] = final KE Because of the way KE is defined, this situation would make the final KE twice as much as the initial KE. So this doe not happen! |
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Calculate the work done in lifting a 500 N barbell 2.2 m above the floor. What is the potential energy of the barbell when it is lifted to this height?
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W= Fd; W= 500N (2.2m); W= 1100 J; PE= 1100 J
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Calculate the watts of power expanded when a force of 1 N moves a book 2 m in a time interval of 0.5 s
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Power=work=1n(2m) =2joules=2watts time 1second second
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How many joules of potential energy does a 1-kg book gain when it is elevated 4 m? When it is elevated 8 m?
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G.P.E. = mgh = 1 kg x 10 m/s/s x 4 m = 40 joules.
G.P.E. = mgh = 1 kg x 10 m/s/s x 8 m = 80 joules |
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Calculate the change in potentail energy 8 million kg of water dropping 50 m over Niagara Falls
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Potential energy(U) = mgh
U=8kgx10^6(9.8m/s^2)(50m) U=3920000000J U=3.92GJ |
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How many joules of kinetic energy does a 1-kg book have when it is tossed across the room at speed of 2m/s?
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KE= ½ m/v^2= ½ 1kg (2 m/s)^2= 2 kg X m^2/ sec^2 = 2 J
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How much work is required to increase the kinetic energy of a car by 5000 N?
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5000 Joules
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What change in kinetic energy does an airplane experience on takeoff if it is moved a distance of 500 m by a sustained net force of 5000 n?
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Energy imparted by the force is converted to change in kinetic energy. The energy from the force = F*d, or 5000*500 J
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The 2nd floor of a house is 6m above street level. How much work is required to lift a 300 kg refrigerfator?
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Work= mgh; 300kg X 10m/s^2 X 6m = 1.8 X 10^4 J
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Which produces the greater change in kinetic energy? Exerting a 10 N force for a distance of 5 m, or exerting a 20 n force over a distance of 2 m?
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10 N X 5m = 50 J; 20 N X 2m = 40 J; Thus the first one produces a greater change
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This question is typical on some driver's license exams: A car moving at 50 km/h skids 16 m with locked brakes. How far will the car skid with locked brakes at 150 km/h?
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F × d = ½ × m × v2; d ∝ v2
- Therefore the stopping distance is proportional to the square of the velocity, and if you are going twice as fast, you require four times the stopping distance! Applying this to our particular problem, the stopping distance at 50 km/h is 15 m. If we increase speed to 150 km/h (three times the initial speed) we expect the stopping distance to rise by three squared, or nine times. Therefore the expected stopping distance is 9 × 15 = 135 m. |
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A 60 kg skydiver moving at terminal speed falls 50 m in 1s. What power is the skydiver expending on the air?
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- power = work/time = force x (distance/time)
=forcexspeed since dist/time=speed so the force is the weight of the skydiver, mg, and the speed is 50m/s so power = mgv = 60kgx9.8m/s/sx50m/s=29400W |
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Consider the inelastic collision between the two freight cars shown here (Figure 3.13). The momentum before and after the collision is the same. The KE, however, is less after the collision than before the collision. How much less, and what has become of this energy?
Remember that the two freight cars have the same mass, which we'll represent as m. The initial kinetic energy is that of the single freight car moving at v = 10 (no units are given, so we won't worry about units): |
½ × m × 102 = 50m. (The momentum is 10m.)
After the collision, the momentum of the total system is the same 10m, but since the moving cars now mass 2m, the final velocity is 5. The kinetic energy is found using these values: ½ × 2m × 52 = 25m, or only half the kinetic energy before the collision!!! What has become of this energy? It has been "dissipated" by the shock of collision, for example in the jolt and rattling experienced by the cars. |
