Work, Energy, and Power

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AP Physics 1 › Work, Energy, and Power

Questions 1 - 10
1

A 50kg man pushes against a wall with a force of 100N for 10 seconds. How much work does the man accomplish?

Explanation

The answer is because no work is done. For work to be done a force must be exerted across a distance parallel to the direction of the force. In this case, a force is exerted by the man but the wall is stationary and since it does not move there is no distance for work to take place on. The formula for work can be written as:

Here, is zero, so is also zero, making the entire term, and thus the work zero.

2

Which of the following variables are involved in the calculation of work?

I. Force

II. Distance

III. Acceleration

I, II and III

I only

II only

I and II

Explanation

Work is defined as:

Where is work, is force experienced by the object/person, and is the distance the object/person moves as a result of the force. Recall that force is calculated as follows.

Where is mass and is acceleration; therefore, work depends on the mass and acceleration (force), and distance.

3

A ball of mass is thrown at a target. The ball strikes with a velocity of and bounces back with equal magnitude. Determine the work done on the ball.

None of these

Explanation

There is no net work done on the ball. The wall stopped the ball, doing negative work, then accelerated the ball, doing positive work. These end up canceling each other out.

4

A ball is initially compressed against a spring on a frictionless horizontal table. The ball is the released, and is shot to the right by the spring. Calculate the work done by the spring on the ball, .

Can not be determined.

Explanation

From the Work-energy theorem,

.

From conservation of energy, we see that all of the potential energy contained in the spring will be transferred into kinetic energy of the ball, shown as

Noting the initial kinetic energy of the ball is , we can see that the work done by the spring will be equal to the initial potential energy of the spring.

5

How much work (in kilojoules) is done to accelerate a car (3000kg) from rest to .

5400 kilojoules

3000 kilojoules

6200 kilojoules

3400 kilojoules

2300 kilojoules

Explanation

Work is found by finding the change in kinetic energy. Since the car started from rest it had no initial kinetic energy.

Divide by 1000 to convert to kilojoules and get 5400 kilojoules.

6

A car is at rest at the bottom of a hill. later, it is at the top of the hill going . Find the net work done.

Explanation

Initially the car is at rest at the bottom of a hill, this velocity and height are zero.

Converting to

Plugging in values:

7

A super ball is dropped from a height of and returns to the same height. Determine the work done by the ground on the ball.

Impossible to determine

Explanation

There will be no work done.

Energy is a scalar quantity. Since it has the same amount of energy in the end as the beginning, no work was done.

Alternatively, it can be modeled as the floor did negative work during the first half of the bounce (when the ball was compressing), and did positive work of the same magnitude and opposite sign in the second half of the bounce. This is due to the direction of the ball's motion flipping and the Force vector of the floor staying the same.

8

A box slides down a inclined plane. If , calculate the magnitude of the work done by friction on the box, . Your answer should only have significant figures.

Explanation

In general,

In order to calculate the work done by a force, we need to find the angle between the force and the displacement through which the object moves. Since the displacement is directed down the incline, then this means the friction force acts up the incline. This means . We must remember the frictional force is represented as

Where is the normal force and is the coefficient of kinetic friction. On an inclined plane, we can show that . Therefore, we can finally write

Plugging in everything, and noting the magnitude is the absolute value of a quantity, we can plug in for our answer:

9

Consider the following system:

Spinning rod with masses at end

Two spherical masses, A and B, are attached to the end of a rigid rod with length l. The rod is attached to a fixed point, p, which is at the midpoint between the masses and is at a height, h, above the ground. The rod spins around the fixed point in a vertical circle that is traced in grey. is the angle at which the rod makes with the horizontal at any given time ( in the figure).

The rod is initially at rest in its horizontal position. How much work would it take to rotate the rod clockwise until it is vertical, at rest, and mass A is at the top?

Neglect air resistance and internal frictional forces. Ignore the mass of the rod itself.

None of these

Explanation

We can use the expression for conservation of energy:

Since the rod is both initially and finally at rest, we can removed both kinetic energies. Also, if we assume point p is at a height of 0, we can removed initial potential energy, leaving us with:

Plugging in the expression for potential energy and expanding for both masses:

Since the rod is vertical, we know that mass A is half a rod's length above our reference height, and mass B is half a rod's length below it. Thus we get:

Factoring to clean up our expression:

We know all of our variables, so time to plug and chug:

10

A ball is initially compressed against a spring on a frictionless horizontal table. The ball is the released, and is shot to the right by the spring. Calculate the work done by the spring on the ball, .

Can not be determined.

Explanation

From the Work-energy theorem,

.

From conservation of energy, we see that all of the potential energy contained in the spring will be transferred into kinetic energy of the ball, shown as

Noting the initial kinetic energy of the ball is , we can see that the work done by the spring will be equal to the initial potential energy of the spring.

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