High School Physics : Calculating Work

Study concepts, example questions & explanations for High School Physics

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Example Questions

Example Question #3 : Work

 box is dropped . How much work was done on the box?

Possible Answers:

Correct answer:

Explanation:

The formula for work is , work equals force times distance.

In this case, there is only one force acting upon the object: the force due to gravity. Plug in our given information for the distance to solve for the work done by gravity.

Remember, since the object will be moving downward, the distance should be negative.

The work done is positive because the distance and the force act in the same direction.

Example Question #4 : Work

 book falls off the top of a  bookshelf. How much work is required to lift the book back to its original position, assuming the lifting is done with a constant velocity?

Possible Answers:

Correct answer:

Explanation:

Work is a force times a distance:

We know the distance that the book needs to travel, but we need to sovle for the lifting force required to move it.

There are two forces acting upon the book: the lifting force and gravity. Since the book is moving with a constant velocity, that means the net force will be zero. Mathemetically, that would look like this:

We can expand the right side of the equation using Newton's second law:

Use the given mass and value of gravity to solve for the lifting force.

Now that we have the force and the distance, we can solve for the work to lift the book.

This problem can also be solved using energy. Work is equal to the change in potential energy:

While on the ground, the book has zero potential energy. Once back on the shelf, the energy is equal to . The work is thus also equal to .

Example Question #5 : Work

 book falls off the top of a  bookshelf. How much work is required to put the book back on the top of the bookshelf, assuming it is lifted with a constant velocity?

Possible Answers:

Correct answer:

Explanation:

Work is a force times a distance:

We know the distance that the book needs to travel, but we need to sovle for the lifting force required to move it.

There are two forces acting upon the book: the lifting force and gravity. Since the book is moving with a constant velocity, that means the net force will be zero. Mathemetically, that would look like this:

We can expand the right side of the equation using Newton's second law:

Use the given mass and value of gravity to solve for the lifting force.

Now that we have the force and the distance, we can solve for the work to lift the book.

This problem can also be solved using energy. Work is equal to the change in potential energy:

While on the ground, the book has zero potential energy. Once back on the shelf, the energy is equal to . The work is thus also equal to .

Example Question #1 : Calculating Work

 book falls off the top of a  bookshelf. How much work is required to put the book back on the top of the bookshelf, assuming it is lifted with a constant velocity?

Possible Answers:

Correct answer:

Explanation:

Work is a force times a distance:

We know the distance that the book needs to travel, but we need to sovle for the lifting force required to move it.

There are two forces acting upon the book: the lifting force and gravity. Since the book is moving with a constant velocity, that means the net force will be zero. Mathemetically, that would look like this:

We can expand the right side of the equation using Newton's second law:

Use the given mass and value of gravity to solve for the lifting force.

Now that we have the force and the distance, we can solve for the work to lift the book.

This problem can also be solved using energy. Work is equal to the change in potential energy:

While on the ground, the book has zero potential energy. Once back on the shelf, the energy is equal to . The work is thus also equal to .

Example Question #402 : High School Physics

A cat knocks a  toy mouse across the floor with  of force. If the cat did  of work, how far did the mouse travel?

Possible Answers:

Correct answer:

Explanation:

The relationship between work, force, and distance is:

We are given the force on the toy and the work done. Using these values, we can find the distance. Note that the mass is not relevant for this question.

Example Question #2 : Calculating Work

A cat knocks a  toy mouse across the floor with  of force. If the toy travels , how much work did the cat do?

Possible Answers:

Correct answer:

Explanation:

The relationship between work, force and distance is:

We are given the value for the force and the distance that the toy travels. Using these values, we can find the work done by the cat. Note that the mass of the toy is not relevant for this calculation.

Example Question #3 : Calculating Work

A cat knocks a  toy mouse across the floor. If it travels  and the cat does  of work, how much force did the cat exert on the mouse?

Possible Answers:

Correct answer:

Explanation:

The relationship between work, force and distance is:

We are given the value for the work done by the cat and the distance that the toy travels. Using these values, we can find the force on the toy. Note that the mass of the toy is not relevant for this calculation.

Example Question #1 : Work

A crate is pushed across the floor. If of force was used to achieve this motion, how much work was done?

Possible Answers:

Correct answer:

Explanation:

The formula for work is:

Given the values for force and distance, we can calculate the work done.

Note that no work is done by the force of gravity or the weight of the box, since the vertical position does not change.

Example Question #4 : Calculating Work

Ryan pushes a  filing cabinet  across the floor, doing  of work. How much force is he applying to the cabinet?

Possible Answers:

Correct answer:

Explanation:

Work is the product of force times a distance:

We are given the work and the distance traveled, allowing us to solve for the force. The mass of the cabinet is not necessary information.

Example Question #5 : Calculating Work

A 1500kg rocket has a net propulsion force of 500N. Over a short period of time, the rocket speeds up uniformly from an initial velocity of  to a final velocity of . Assume that the mass of the rocket remains constant even though it is burning fuel and that the net force is along the direction of travel. What is the net work done on the rocket in kJ?

Possible Answers:

None of these

Correct answer:

Explanation:

Use the data given to calculate the kinetic energy of the rocket at the two different velocities. Then find the amount of work done using the following equation: 

Kinetic energy of the rocket at the two velocities:

The change in the kinetic energy at the two velocities:

 

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