Work - High School Physics
Card 0 of 20
A 
box is dropped 
. How much work was done on the box?
A
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.
The formula for work is
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.
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A 
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?
A
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 
.
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
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A 
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?
A
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 
.
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
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A 
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?
A
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 
.
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
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A cat knocks a 
toy mouse across the floor with 
of force. If the cat did 
of work, how far did the mouse travel?
A cat knocks a
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.
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.
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A cat knocks a 
toy mouse across the floor with 
of force. If the toy travels 
, how much work did the cat do?
A cat knocks a
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.
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.
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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?
A cat knocks a
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.
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.
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A 
crate is pushed 
across the floor. If 
of force was used to achieve this motion, how much work was done?
A
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.
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.
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Ryan pushes a 
filing cabinet 
across the floor, doing 
of work. How much force is he applying to the cabinet?
Ryan pushes a
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.
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.
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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?
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
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:
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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A bodybuilder lifts a 
weight upward 
, and then returns it to its original position, all with a constant speed. What is the net work done on the weight?
A bodybuilder lifts a
Work is a vector quantity equal to the change in energy in a given direction. Our net work will be equal to product of force and displacement.
We can write this equation in terms of Newton's second law to incorporate our given values.
During its motion the weight travels a distance of 
upward, and then 
downward. Its total distance will be 
, however, we are looking for its displacement. The initial height and final height are equal, making the displacement equal to zero.
The total work will be zero because there was no net displacement.
Work is a vector quantity equal to the change in energy in a given direction. Our net work will be equal to product of force and displacement.
We can write this equation in terms of Newton's second law to incorporate our given values.
During its motion the weight travels a distance of
The total work will be zero because there was no net displacement.
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A 
crate slides along the floor for 
. What other piece of information do we need to find the work done?
A
The formula for work is 
, or work equals force times distance.
Since 
, you can expand the work formula a little bit to see 
.
The problem gives us the mass and the distance traveled, but not the acceleration. Given the acceleration, along with the mass and distance, we would be able to calculate the work.
The formula for work is
Since
The problem gives us the mass and the distance traveled, but not the acceleration. Given the acceleration, along with the mass and distance, we would be able to calculate the work.
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Miguel pushes a crate with 
of force. What other piece of information do we need to find the work he does?
Miguel pushes a crate with
The formula for work is 
, or work equals force times distance.
The problem gives us the mass, but is missing the distance. Given the distance, we would be able to calculate the work.
The formula for work is
The problem gives us the mass, but is missing the distance. Given the distance, we would be able to calculate the work.
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Which of the following is an example of work?
Which of the following is an example of work?
Work is a force times a distance or 
.
Only one of these has a force times a distance: the weightlifter lifting a weight above his head.
The car moving at a constant velocity has no acceleration and therefore no force.
Typing essays does not require a force times a distance (though it sure feels like work!).
If the skier has no loss in energy, then no work was done, as work is also the measurement of the change in energy.
The man sitting in a chair has a constant velocity of zero, no acceleration, and therefore no force.
Work is a force times a distance or
Only one of these has a force times a distance: the weightlifter lifting a weight above his head.
The car moving at a constant velocity has no acceleration and therefore no force.
Typing essays does not require a force times a distance (though it sure feels like work!).
If the skier has no loss in energy, then no work was done, as work is also the measurement of the change in energy.
The man sitting in a chair has a constant velocity of zero, no acceleration, and therefore no force.
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Which of these is a correct formula for work?
Which of these is a correct formula for work?
Work is produced when a force is used to create a change in distance. The amount of work is given by the product of the distance traveled and the force applied.
Work is produced when a force is used to create a change in distance. The amount of work is given by the product of the distance traveled and the force applied.
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Michael pushes a box 
across the floor. If the box moves with a constant velocity, what is the total work done?
Michael pushes a box
Works is equal to force times distance:
Since the box moves with a constant velocity, it has no acceleration. Remember that 
, so if there is no acceleration, then there must be no net force.
If there is no net force, then there is also no work.
Works is equal to force times distance:
Since the box moves with a constant velocity, it has no acceleration. Remember that
If there is no net force, then there is also no work.
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Rodrigo pushes a sofa 
down a long hallway. If the couch moves with a constant velocity, how much work does Rodrigo do?
Rodrigo pushes a sofa
Work is the product of a net force over a given distance.
In order for there to be work, there must be a net force applied and the object must have a non-zero displacement. In this question we are given the displacement, but we have to solve for the force.
The question only tells us that the couch moves with constant velocity. This tells us that acceleration is zero. If acceleration is zero, then force must also be zero according to Newton's second law.
If the force is zero, then the work is also zero.
Work is the product of a net force over a given distance.
In order for there to be work, there must be a net force applied and the object must have a non-zero displacement. In this question we are given the displacement, but we have to solve for the force.
The question only tells us that the couch moves with constant velocity. This tells us that acceleration is zero. If acceleration is zero, then force must also be zero according to Newton's second law.
If the force is zero, then the work is also zero.
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A 
arrow is fired from a bow whose string exerts an average force of 
on the arrow over a distance of 
. What is the speed of the arrow as it leaves the bow?
A 
Known
Unknown
We can use the work kinetic energy theorem to solve this problem. The work done to the object causes a change in kinetic energy.
Work is equal to the force times the displacement over which the force acted.
Therefore
Plug in our variables and solve
Known
Unknown
We can use the work kinetic energy theorem to solve this problem. The work done to the object causes a change in kinetic energy.
Work is equal to the force times the displacement over which the force acted.
Therefore
Plug in our variables and solve
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What is the work applied to a 
car when it is pushed 
up along a 
incline at a constant velocity.. Assume the coefficient of friction between the car and the ramp is 
.
What is the work applied to a 
Known
Unknown
The work done on the object is equal to the net force done on the object multiplied by the displacement through which the force acts.
Therefore the first thing to do in this problem is to determine the amount of force needed to push the car up the incline. Since the car is moving at a constant velocity, all forces must be balanced. The force in the direction of motion is the force applied which is doing work on the object. The forces opposing motion are the force of friction and the x-component of the gravitational force.
To determine the Force of Gravity in the x-direction, we must break the force of gravity into components and examine the side acting in the x-direction. Using trigonometric functions we get that
We know that the force of gravity is equal to mg
The force of friction is directly related to μ (the coefficient of friction) times the normal force. In this case the normal force is equal to the y component of the force of gravity.
Therefore
We know that our force applied must balance both of these forces.
We can now substitute this equation into our work each to determine the work applied to the car.
We can now substitute in our variables and solve for the missing piece.
Known
Unknown
The work done on the object is equal to the net force done on the object multiplied by the displacement through which the force acts.
Therefore the first thing to do in this problem is to determine the amount of force needed to push the car up the incline. Since the car is moving at a constant velocity, all forces must be balanced. The force in the direction of motion is the force applied which is doing work on the object. The forces opposing motion are the force of friction and the x-component of the gravitational force.
To determine the Force of Gravity in the x-direction, we must break the force of gravity into components and examine the side acting in the x-direction. Using trigonometric functions we get that
We know that the force of gravity is equal to mg
The force of friction is directly related to μ (the coefficient of friction) times the normal force. In this case the normal force is equal to the y component of the force of gravity.
Therefore
We know that our force applied must balance both of these forces.
We can now substitute this equation into our work each to determine the work applied to the car.
We can now substitute in our variables and solve for the missing piece.
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At an accident science on a level road, investigators measure a car’s skid mark to be 
long. The accident occurred on a rainy day, the coefficient of friction was estimated to be 
. What was the speed of the car when the driver slammed on and locked the brakes?
At an accident science on a level road, investigators measure a car’s skid mark to be 
Knowns
Unknown
We know that the force of friction is what caused the skid marks and is the cause of the work done on car to slow it to a stop. We also know that the work is equal to the change in kinetic energy.
Work is equal to the force times the displacement over which the force acted.
Therefore
The force in this case is the force of friction.
Since the car is on a level road the normal force is equal and opposite the force of gravity.
The force of gravity is equal to the mass times the acceleration due to gravity
Therefore
We can substitute this in for our work-kinetic energy equation.
Since the final velocity is 
we can cancel this out of the problem.
Since there is a negative and a mass in both terms we can cancel these out of the equation
Now we can substitute in our values and solve.
Knowns
Unknown
We know that the force of friction is what caused the skid marks and is the cause of the work done on car to slow it to a stop. We also know that the work is equal to the change in kinetic energy.
Work is equal to the force times the displacement over which the force acted.
Therefore
The force in this case is the force of friction.
Since the car is on a level road the normal force is equal and opposite the force of gravity.
The force of gravity is equal to the mass times the acceleration due to gravity
Therefore
We can substitute this in for our work-kinetic energy equation.
Since the final velocity is 
Since there is a negative and a mass in both terms we can cancel these out of the equation
Now we can substitute in our values and solve.
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