PE & KE Quiz

1. As a spring is stretched, its elastic potential energy

Decreases

Increases

Remains the same

When a spring is stretched, its elastic potential energy increases. This is because the spring is being deformed and work is being done to stretch it. The potential energy is stored in the spring and can be released when the spring returns to its original position. Therefore, as the spring is stretched further, the amount of potential energy stored in it increases.

Explanation

When a spring is stretched, its elastic potential energy increases. This is because the spring is being deformed and work is being done to stretch it. The potential energy is stored in the spring and can be released when the spring returns to its original position. Therefore, as the spring is stretched further, the amount of potential energy stored in it increases.

2. Using ground level as the reference height with zero potential energy, which object has the greatest gravitational potential energy?

2-kg mass at 60-m height

5-kg mass at 5-m height

20-kg mass at 50-m height

40-kg mass at 2-m height

The gravitational potential energy of an object depends on its mass and height. The formula for gravitational potential energy is mass multiplied by height multiplied by the acceleration due to gravity. Comparing the given options, the 20-kg mass at 50-m height has the greatest gravitational potential energy because it has the highest product of mass and height compared to the other options. The 2-kg mass at 60-m height has a lower product, the 5-kg mass at 5-m height has an even lower product, and the 40-kg mass at 2-m height has the lowest product among the options. Therefore, the 20-kg mass at 50-m height has the greatest gravitational potential energy.

Explanation

The gravitational potential energy of an object depends on its mass and height. The formula for gravitational potential energy is mass multiplied by height multiplied by the acceleration due to gravity. Comparing the given options, the 20-kg mass at 50-m height has the greatest gravitational potential energy because it has the highest product of mass and height compared to the other options. The 2-kg mass at 60-m height has a lower product, the 5-kg mass at 5-m height has an even lower product, and the 40-kg mass at 2-m height has the lowest product among the options. Therefore, the 20-kg mass at 50-m height has the greatest gravitational potential energy.

3. A city's water tower has a capacity of 1.0 E 3kg of water. A pump is filling the water tower to an average height of 50.0 m.

What is the potential energy of the water in the tower?

2.5 E 4 J

5.0 E 4 J

4.9 E 5 J

2.5 E 6 J

The potential energy of an object is given by the equation PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. In this case, the mass of the water is given as 1.0 E 3 kg and the height is given as 50.0 m. Plugging these values into the equation, we get PE = (1.0 E 3 kg)(9.8 m/s^2)(50.0 m) = 4.9 E 5 J. Therefore, the potential energy of the water in the tower is 4.9 E 5 J.

Explanation

The potential energy of an object is given by the equation PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. In this case, the mass of the water is given as 1.0 E 3 kg and the height is given as 50.0 m. Plugging these values into the equation, we get PE = (1.0 E 3 kg)(9.8 m/s^2)(50.0 m) = 4.9 E 5 J. Therefore, the potential energy of the water in the tower is 4.9 E 5 J.

4. How high can a worker lift a 40.0-kg bag of sand if he produces 4,000.0 J of energy? Assume no energy is used to overcome friction.

1.020 m

10.20 m

102.0 m

1,020 m

The worker can lift the 40.0-kg bag of sand to a height of 10.20 m because the amount of energy produced is directly proportional to the height the object can be lifted. The formula for potential energy is PE = mgh, where PE is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height. Rearranging the formula to solve for h, we get h = PE / (mg). Plugging in the given values, we get h = 4000 / (40 * 9.8) = 10.20 m.

Explanation

The worker can lift the 40.0-kg bag of sand to a height of 10.20 m because the amount of energy produced is directly proportional to the height the object can be lifted. The formula for potential energy is PE = mgh, where PE is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height. Rearranging the formula to solve for h, we get h = PE / (mg). Plugging in the given values, we get h = 4000 / (40 * 9.8) = 10.20 m.

5. A 12.5-kg glider is observed flying at an altitude of 1,510 m at a constant velocity of 18.0 m/s. The glider dives to a new altitude of 1,250 m

Neglecting the effects of air resistance, what is its change in potential energy?

31,882 J

153,000 J

85,000 J

338,000 J

The change in potential energy of the glider as it moves from an initial altitude (h1) to a final altitude (h2) can be calculated using the following formula:

ΔPE = m * g * (h2 - h1)

Where:

ΔPE = Change in potential energy

m = Mass of the glider (12.5 kg)

g = Acceleration due to gravity (approximately 9.81 m/s²)

h2 = Final altitude (1,250 m)

h1 = Initial altitude (1,510 m)

Now, let's plug in the values and calculate:

ΔPE = 12.5 kg * 9.81 m/s² * (1,250 m - 1,510 m)

ΔPE = 12.5 kg * 9.81 m/s² * (-260 m)

ΔPE = -31,882 J

The change in potential energy is -31,882 joules (J). The negative sign indicates that the glider's potential energy has decreased as it moved to a lower altitude.

Explanation

The change in potential energy of the glider as it moves from an initial altitude (h1) to a final altitude (h2) can be calculated using the following formula:

ΔPE = m * g * (h2 - h1)

Where:

ΔPE = Change in potential energy

m = Mass of the glider (12.5 kg)

g = Acceleration due to gravity (approximately 9.81 m/s²)

h2 = Final altitude (1,250 m)

h1 = Initial altitude (1,510 m)

Now, let's plug in the values and calculate:

ΔPE = 12.5 kg * 9.81 m/s² * (1,250 m - 1,510 m)

ΔPE = 12.5 kg * 9.81 m/s² * (-260 m)

ΔPE = -31,882 J

The change in potential energy is -31,882 joules (J). The negative sign indicates that the glider's potential energy has decreased as it moved to a lower altitude.

6. A 1.00-kilogram ball is dropped from the top of a building. Just before striking the ground, the ball's speed is 12.0 meters per second. What was the ball's gravitational potential energy, relative to the ground, at the instant it was dropped? [Neglect friction.]

6.00 J

72.0 J

24.0 J

144 J

When the ball is dropped from the top of the building, its initial potential energy is converted into kinetic energy as it falls. The gravitational potential energy of an object is given by the formula PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object. In this case, the mass of the ball is 1.00 kg and the height is not given. However, since the ball is just about to strike the ground, we can assume that the height is negligible. Therefore, the gravitational potential energy of the ball at the instant it was dropped is approximately 72.0 J.

Explanation

When the ball is dropped from the top of the building, its initial potential energy is converted into kinetic energy as it falls. The gravitational potential energy of an object is given by the formula PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object. In this case, the mass of the ball is 1.00 kg and the height is not given. However, since the ball is just about to strike the ground, we can assume that the height is negligible. Therefore, the gravitational potential energy of the ball at the instant it was dropped is approximately 72.0 J.

7. A father (100 kg) and his son (50 kg) are jogging at the same speed. Which statement is true about the kinetic energies (KE) of the father and the son?

KE of father = 2KE of son

KE of father = 1/2KE of son

KEof father = 4KE of son

KE of father = 1/4KE of son

The kinetic energy of an object is given by the equation KE = 1/2mv^2, where m is the mass of the object and v is its velocity. Since the father and the son are jogging at the same speed, their velocities are the same. However, the father has twice the mass of the son. Therefore, the kinetic energy of the father (KE of father) would be twice that of the son (2KE of son).

Explanation

The kinetic energy of an object is given by the equation KE = 1/2mv^2, where m is the mass of the object and v is its velocity. Since the father and the son are jogging at the same speed, their velocities are the same. However, the father has twice the mass of the son. Therefore, the kinetic energy of the father (KE of father) would be twice that of the son (2KE of son).

8. 14. Which graph best represents the relationship between the gravitational potential energy of an object near the surface of Earth and its height above Earth's surface?

1

2

3

4

The correct answer is 1. This graph represents the relationship between gravitational potential energy and height above Earth's surface. As the height increases, the gravitational potential energy also increases. This is because the gravitational potential energy is directly proportional to the height. The graph shows a linear relationship between the two variables, with a positive slope indicating the direct proportionality.

Explanation

The correct answer is 1. This graph represents the relationship between gravitational potential energy and height above Earth's surface. As the height increases, the gravitational potential energy also increases. This is because the gravitational potential energy is directly proportional to the height. The graph shows a linear relationship between the two variables, with a positive slope indicating the direct proportionality.

9. A skateboarder with a mass of 50. kg is riding on a half-pipe as shown in the diagram below. He has a speed of 5.0 m/s at the bottom.

What vertical distance will the skateboarder climb?

0.25 m

0.50 m

1.3 m

2.5 m

The skateboarder will climb a vertical distance of 1.3 m. This can be determined by using the principle of conservation of mechanical energy. At the bottom of the half-pipe, the skateboarder has kinetic energy due to his speed. As he climbs up, this kinetic energy is converted into potential energy. The potential energy is given by the equation PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the vertical distance. By equating the initial kinetic energy to the final potential energy, we can solve for h and find that it is equal to 1.3 m.

Explanation

The skateboarder will climb a vertical distance of 1.3 m. This can be determined by using the principle of conservation of mechanical energy. At the bottom of the half-pipe, the skateboarder has kinetic energy due to his speed. As he climbs up, this kinetic energy is converted into potential energy. The potential energy is given by the equation PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the vertical distance. By equating the initial kinetic energy to the final potential energy, we can solve for h and find that it is equal to 1.3 m.

10. Which graph best represents the relationship between the kinetic energy of a moving object and its velocity?

1

2

3

4

Graph 3 best represents the relationship between the kinetic energy of a moving object and its velocity because as the velocity increases, the kinetic energy also increases. This is shown by the upward slope of the line in graph 3. Graphs 1 and 2 show a constant kinetic energy regardless of velocity, which is incorrect. Graph 4 shows a decreasing kinetic energy with increasing velocity, which is also incorrect. Therefore, graph 3 is the best representation of the relationship between kinetic energy and velocity.

Explanation

Graph 3 best represents the relationship between the kinetic energy of a moving object and its velocity because as the velocity increases, the kinetic energy also increases. This is shown by the upward slope of the line in graph 3. Graphs 1 and 2 show a constant kinetic energy regardless of velocity, which is incorrect. Graph 4 shows a decreasing kinetic energy with increasing velocity, which is also incorrect. Therefore, graph 3 is the best representation of the relationship between kinetic energy and velocity.

11. In the diagram below, a wooden block slides from rest down a frictionless incline. The block attains a speed of 3 m/s at the bottom of the incline.

How high is the incline?

0.21 m

0.31 m

0.46 m

0.59 m

The height of the incline can be determined by using the principle of conservation of energy. As the block slides down the incline, it gains kinetic energy. This energy comes from the potential energy it had at the top of the incline. The potential energy can be calculated using the equation PE = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height of the incline. The kinetic energy can be calculated using the equation KE = 0.5mv^2, where m is the mass of the block and v is the speed of the block at the bottom of the incline. Since the block is initially at rest, all of its initial potential energy is converted into kinetic energy. Therefore, we can equate the two equations: mgh = 0.5mv^2. The mass cancels out, and we can solve for h: h = 0.5v^2/g = 0.5(3^2)/9.8 = 0.46 m.

Explanation

The height of the incline can be determined by using the principle of conservation of energy. As the block slides down the incline, it gains kinetic energy. This energy comes from the potential energy it had at the top of the incline. The potential energy can be calculated using the equation PE = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height of the incline. The kinetic energy can be calculated using the equation KE = 0.5mv^2, where m is the mass of the block and v is the speed of the block at the bottom of the incline. Since the block is initially at rest, all of its initial potential energy is converted into kinetic energy. Therefore, we can equate the two equations: mgh = 0.5mv^2. The mass cancels out, and we can solve for h: h = 0.5v^2/g = 0.5(3^2)/9.8 = 0.46 m.

12. A 1.0-kilogram rubber ball traveling east at 4.0 meters per second hits a wall and bounces back toward the west at 2.0 meters per second. Compared to the kinetic energy of the ball before it hits the wall, the kinetic energy of the ball after it bounces off the wall is

One-fourth as great

The same

One-half as great

Four times as great

When the rubber ball hits the wall, it undergoes a change in velocity. The initial velocity of the ball is 4.0 meters per second towards the east, and after bouncing off the wall, the velocity becomes 2.0 meters per second towards the west. Kinetic energy is directly proportional to the square of velocity. Since the velocity is halved (2.0 divided by 4.0) after the bounce, the kinetic energy will be one-fourth (1/2 squared) as great as before the bounce. Therefore, the kinetic energy of the ball after bouncing off the wall is one-fourth as great as before it hits the wall.

Explanation

When the rubber ball hits the wall, it undergoes a change in velocity. The initial velocity of the ball is 4.0 meters per second towards the east, and after bouncing off the wall, the velocity becomes 2.0 meters per second towards the west. Kinetic energy is directly proportional to the square of velocity. Since the velocity is halved (2.0 divided by 4.0) after the bounce, the kinetic energy will be one-fourth (1/2 squared) as great as before the bounce. Therefore, the kinetic energy of the ball after bouncing off the wall is one-fourth as great as before it hits the wall.

13. The graph below shows the relationship between the elongation of a spring and the force applied to the spring causing it to stretch. What is the spring constant for this spring?

0.020 N/m

25 N/m

2.0 N/m

50 N/m

The spring constant for a spring is a measure of its stiffness, or how much force is required to stretch or compress it. In this graph, the relationship between elongation and force is linear, meaning that as the force applied to the spring increases, the elongation also increases at a constant rate. The slope of the line on the graph represents the spring constant. In this case, the slope is steep, indicating a large spring constant. Therefore, the correct answer is 50 N/m.

Explanation

The spring constant for a spring is a measure of its stiffness, or how much force is required to stretch or compress it. In this graph, the relationship between elongation and force is linear, meaning that as the force applied to the spring increases, the elongation also increases at a constant rate. The slope of the line on the graph represents the spring constant. In this case, the slope is steep, indicating a large spring constant. Therefore, the correct answer is 50 N/m.

14. The diagram below represents the path of Earth around the Sun.

As Earth travels in its orbit from its January position to its July position, the potential energy of Earth

Decreases and its kinetic energy decreases

Decreases and its kinetic energy increases

Increases and its kinetic energy decreases

Increases and its kinetic energy increases

As Earth travels from its January position to its July position, it moves closer to the Sun. This decrease in distance results in a decrease in potential energy, as the gravitational force between Earth and the Sun becomes stronger. At the same time, the increase in speed as Earth moves closer to the Sun leads to an increase in kinetic energy. Therefore, the potential energy decreases and the kinetic energy increases.

Explanation

As Earth travels from its January position to its July position, it moves closer to the Sun. This decrease in distance results in a decrease in potential energy, as the gravitational force between Earth and the Sun becomes stronger. At the same time, the increase in speed as Earth moves closer to the Sun leads to an increase in kinetic energy. Therefore, the potential energy decreases and the kinetic energy increases.

15. A catapult with a spring constant of 1.0 m 10⁴ newtons per meter is required to launch an airplane from the deck of an aircraft carrier. The plane is released when it has been displaced 0.50 meter from its equilibrium position by the catapult. The energy acquired by the airplane from the catapult during takeoff is approximately

1.3 E 3J

2.5 E 3 J

2.0 E 4J

1.0 E 4 J

The energy acquired by the airplane from the catapult during takeoff can be calculated using the formula for potential energy stored in a spring, which is given by 1/2 * k * x^2, where k is the spring constant and x is the displacement from the equilibrium position. Plugging in the values, we get 1/2 * (1.0 * 10^4 N/m) * (0.50 m)^2 = 1.25 * 10^3 J, which is approximately 1.3 E 3J.

Explanation

The energy acquired by the airplane from the catapult during takeoff can be calculated using the formula for potential energy stored in a spring, which is given by 1/2 * k * x^2, where k is the spring constant and x is the displacement from the equilibrium position. Plugging in the values, we get 1/2 * (1.0 * 10^4 N/m) * (0.50 m)^2 = 1.25 * 10^3 J, which is approximately 1.3 E 3J.