Chpt. 7 HW

In the current window, click and drag a new track (the shape with three circles in the bottom left of the window), and place it near the upper left end of the existing track until the two connect. Then, click and drag on the circles to stretch and/or bend the track to make it look as shown below. If the skater starts from rest at position 1, rank, in increasing order from least to greatest, the kinetic energy of the skater at the five positions shown.

1, 3, 2&5, 4

One common application of conservation of energy in mechanics is to determine the speed of an object. Although the simulation doesn't give the skater's speed, you can calculate it because the skater's kinetic energy is known at any location on the track. Consider again the case where the skater starts 7 m above the ground and skates down the track. What is the skater's speed when the skater is at the bottom of the track? Express your answer numerically in meters per second to two significant figures.

11 m/s For a 75-kg object having approximately 4410 J of energy, the speed must be roughly 11 m/s.

Ignoring friction, the total energy of the skater is conserved. This means that the kinetic plus potential energy at one location, say E1=K1+U1, must be equal to the kinetic plus potential energy at a different location, say E2=K2+U2. This is the principle of conservation of energy and can be expressed as E1=E2. Since the energy is conserved, the change in the kinetic energy is equal to the negative of the change in the potential energy: K2−K1=−(U2−U1), or ΔK1=−ΔU2. At the bottom of the simulation window, click on Playground. For this part of the tutorial, you should have the Friction slider (in the upper right part of the window) set to "None," which means no thermal energy is generated. Select the Show Grid option. Then, add a track by clicking and dragging on a new track (the shape with three circles in the bottom left of the window) and placing it near the skater. You can then click and drag on individual circles to stretch and /or bend the track and make it look as shown below. The bottom of the track should be 1 m above the ground, and both ends of the track should be at a height of 7 m Place the skater on the track 7 m above the ground, and look at the resulting motion and the Bar Graph showing the energetics. Match the approximate numerical values on the left with the energy type categories on the right to complete the equations. Assume that the mass of the skater is 75.0 kg and that the acceleration of gravity is 9.8 N/kg .

1) Total Energy at Initial Position= 5145 J 2) Potential Energy at Initial Position= 5145 J 3)Kinetic Energy at Initial Position= 0 J 4) Total Energy at Bottom of Track= 5145 J 5) Potential Energy at Bottom of Track= 735 J 6) Kinetic Energy of Bottom of Track= 4410 J

Find the amount of energy E dissipated by friction by the time the block stops.

E = 12mv2+mgh

Rank speed from greatest to least at each point. a) Rank these quantities from greatest to least at each point. b) Rank KE from greatest to least at each point. c) Rank PE from greatest to least at each point.

a) D, B, C, E, A b) D, B, C, E, A c) A, E, C, B, D

What will the kinetic energy of a pile driver ram be if it starts from rest and undergoes a 10 kJ decrease in potential energy? a) -10 kJ b) 10 kJ c) 0 kJ d) 5 kJ

b) 10 kJ

Which requires more work: lifting a 50-kg sack a vertical distance of 2 m or lifting a 25-kg sack a vertical distance of 4 m? a) The sack lifted 4 m requires more work. b) Both take the same 1000 J. c) The 25 kg sack requires more work. d) The 50 kg sack requires more work.

b) Both take the same 1000 J.

Based on the previous question, which statement is true? The kinetic energy at the bottom of the ramp is Based on the previous question, which statement is true? a) equal to the total energy. b) equal to the amount of potential energy loss in going from the initial location to the bottom. c) equal to the initial potential energy.

b) equal to the amount of potential energy loss in going from the initial location to the bottom. Because the total energy is conserved, the kinetic energy at the bottom of the hill plus the potential energy at the bottom of the hill must equal the initial potential energy (since the initial kinetic energy is zero): Kbottom+Ubottom=Uinitial. Solving for the kinetic energy, we get Kbottom=Uinitial−Ubottom, or Kbottom=5145 J−735 J=4410 J. More generally, the change in the kinetic energy is equal to the negative of the change in the potential energy.

Fossil fuels, hydroelectric power, and wind power ultimately get their energy from _______. a) greenhouse gases b) the Sun c) potential energy d) Earth's nuclear energy.

b) the Sun

During a certain time interval, the net work done on an object is zero joules. We can be certain that ____. a) the object was at rest during this entire interval b) the object's final speed was the same as its initial speed c) the object was at rest at the end of this interval

b) the object's final speed was the same as its initial speed When the net work done on an object is zero, there is no overall change in the object's kinetic energy.

Because we are ignoring friction, no thermal energy is generated and the total energy is the mechanical energy, the kinetic energy plus the potential energy: E=K+U. Observe the total energy bar on the Bar Graph. As the skater is skating back and forth, which statement best describes the total energy? The total energy is Because we are ignoring friction, no thermal energy is generated and the total energy is the mechanical energy, the kinetic energy plus the potential energy: . Observe the total energy bar on the Bar Graph. As the skater is skating back and forth, which statement best describes the total energy? a) greatest at the locations where the skater turns and goes back in the opposite direction and smallest at the lowest point of the track. b) the same at all locations of the track. c) smallest at the locations where the skater turns to go back in the opposite direction and greatest at the lowest point of the track.

b) the same at all locations of the track. The mechanical energy (kinetic plus potential) is conserved. (Since there is no friction, the mechanical energy is equal to the total energy.) When the kinetic energy is relatively small, the potential energy is relatively large, and vice versa.

Which most simplified form of the law of conservation of energy describes the motion of the block as it slides on the floor from the bottom of the ramp to the moment it stops? a) 12mv2i+mghi+Wnc=12mv2f+mghf b) 12mv2i=12mv2f c) 12mv2i+Wnc=0 d) 12mv2i+mghi=12mv2f+mghf e) 12mv2i+mghi+12kx2i+Wnc=12mv2f+mghf+12kx2f

c) 12mv2i+Wnc=0

If the skater started from rest 4 m above the ground (instead of 7m), what would be the kinetic energy at the bottom of the ramp (which is still 1 m above the ground)? a) 4410 J b) 735 J c) 2205 J d) 2940 J

c) 2205 J Notice that the skater only drops down 3 m, as compared to 6 m for the case when the skater started 7 m above the ground. In this case, the skater loses half as much potential energy as in the previous case, so the kinetic energy at the bottom is only half as much as the previous case.

When the useful energy output of a simple machine is 100 J, and the total energy input is 200 J, the efficiency is _______. a) 200 % b) 75 %. c) 50 %. d) 100 %

c) 50 %.

As the block slides across the floor, what happens to its kinetic energy K, potential energy U, and total mechanical energy E? a) K decreases;U increases;E decreases b) K increases;U decreases;E decreases c) K decreases;U stays the same;E decreases d) K increases;U stays the same;E decreases e) K decreases;U increases;E stays the same f) K increases;U decreases;E stays the same

c) K decreases;U stays the same;E decreases

Now observe the potential energy bar on the Bar Graph. As the skater is skating back and forth, where does the skater have the most potential energy? The skater's potential energy is Now observe the potential energy bar on the Bar Graph. As the skater is skating back and forth, where does the skater have the most potential energy? a) at its maximum value at the lowest point of the track. b) the same everywhere. c) at its maximum value at the locations where the skater turns and goes back in the opposite direction.

c) at its maximum value at the locations where the skater turns and goes back in the opposite direction. The gravitational potential energy of an object is given by U=mgy, where y is the object's height above the potential energy reference, which is currently the ground. Thus, the skater's potential energy is greatest at the locations where the skater turns to go back in the opposite direction, where the skater is the highest above the reference line. Notice that the skater's potential energy is greatest where the kinetic energy is the lowest, and vice versa.

Where on the track is the skater's kinetic energy the greatest? The skater's kinetic energy is Click on Bar Graph, and observe the kinetic energy bar as the skater goes back and forth. You can select Slow Motion below the track for a more accurate observation. Where on the track is the skater's kinetic energy the greatest? a) at its maximum value at the locations where the skater turns and goes back in the opposite direction. b) the same everywhere. c) at its maximum value at the lowest point of the track.

c) at its maximum value at the lowest point of the track. The kinetic energy of an object is given by K=(1/2)mv2, where v is the speed of the object and m is the mass of the object. Thus, the skater's kinetic energy is greatest at the lowest point of the track, where the skater is moving the fastest.

What force is responsible for the decrease in the mechanical energy of the block? a) tension b) gravity c) friction d) normal force

c) friction

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? a) Immediately before hitting the ground the apple's energy is kinetic energy; when it hits the ground, its energy becomes potential energy. b) Immediately before hitting the ground the apple's energy is thermal energy; when it hits the ground, its energy becomes kinetic energy. c) Immediately before hitting the ground the apple's energy is potential energy; when it hits the ground, its energy becomes thermal energy. d) Immediately before hitting the ground the apple's energy is kinetic energy; when it hits the ground, its energy becomes thermal energy.

d) Immediately before hitting the ground the apple's energy is kinetic energy; when it hits the ground, its energy becomes thermal energy.

Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g=30 m/s2. When he releases the ball from chin height without giving it a push, how will the ball's behavior differ from its behavior on Earth? Ignore friction and air resistance. (Select all that apply.) a) It will smash his face. b) It will take more time to return to the point from which it was released. c) Its mass will be greater. d) It will take less time to return to the point from which it was released. e) It will stop well short of his face.

d) It will take less time to return to the point from which it was released. The pendulum will swing back and forth more quickly (with a shorter period) because it is oscillating in a stronger gravitational field than that on Earth.

When the skater starts 7 m above the ground, how does the speed of the skater at the bottom of the track compare to the speed of the skater at the bottom when the skater starts 4 m above the ground? The speed is When the skater starts 7 above the ground, how does the speed of the skater at the bottom of the track compare to the speed of the skater at the bottom when the skater starts 4 above the ground? a) twice as fast. b) four times as fast. c) the same. d) higher, but less than twice as fast.

d) higher, but less than twice as fast. The person will have twice as much kinetic energy. Because kinetic energy is proportional to the speed squared, the ratio of the speeds is equal to the square root of the ratio of the kinetic energies. In this case, since the ratio of the kinetic energies is 2, the ratio of the speeds is equal to the square root of 2, or roughly 1.4.

The amount of kinetic energy an object has depends on its mass and its speed. Rank the following sets of oranges and cantaloupes from least kinetic energy to greatest kinetic energy. If two sets have the same amount of kinetic energy, place one on top of the other. mass 4m speed 1/4v, total mass 2m speed 1/2v, mass m speed v, total mass 4m speed v = mass 4m speed v

mass 4m speed 1/4v total mass 2m speed 1/2v mass m speed v total mass 4m speed v = mass 4m speed v

Using conservation of energy, find the speed vb of the block at the bottom of the ramp.

vb = (v2)+2gh)