Work - AP Physics 1
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A train of mass 
goes from 
to 
in 
. Calculate the magnitude of force from the brakes.
A train of mass
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Use work:
All energy will be kinetic.
Convert 
to 
:
Plug in values. Force will be negative as it is directed against the direction of travel:
Solve for 
:
Use work:
All energy will be kinetic.
Convert
Plug in values. Force will be negative as it is directed against the direction of travel:
Solve for
A weightlifter doing a bench press lifts a 
weight from the base of his chest to a height that is 
above his chest. How much work has he done after a single lift?
A weightlifter doing a bench press lifts a
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Work is equal to force multiplied by the distance traveled in the direction that the force is applied. The force in this problem is equal to the gravitational force that is created by the weight because that is the minimum force that the weightlifter must apply in order to move the weight. The distance is the distance from the weightlifters chest to the peak of his lift.
Work is equal to force multiplied by the distance traveled in the direction that the force is applied. The force in this problem is equal to the gravitational force that is created by the weight because that is the minimum force that the weightlifter must apply in order to move the weight. The distance is the distance from the weightlifters chest to the peak of his lift.
An 500kg elevator is at rest. If it is raised 50 meters and returns to rest, how much total work was done on the elevator?
An 500kg elevator is at rest. If it is raised 50 meters and returns to rest, how much total work was done on the elevator?
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This can be a tricky question. You need to rely on the work-energy theorem, which states:
Since the elevator is at rest at both the beginning and end, the net work is 0; there is no net change in energy, and therefore no work.
This theorem can be confusing to some since it completely negates potential energy. However, let's think about the situation presented in the problem. A force is required to raise the elevator, meaning that energy is put into the system. However, since it comes back to rest, all of the energy that was put in has been removed by the force of gravity, resulting in a net of zero work.
This can be a tricky question. You need to rely on the work-energy theorem, which states:
Since the elevator is at rest at both the beginning and end, the net work is 0; there is no net change in energy, and therefore no work.
This theorem can be confusing to some since it completely negates potential energy. However, let's think about the situation presented in the problem. A force is required to raise the elevator, meaning that energy is put into the system. However, since it comes back to rest, all of the energy that was put in has been removed by the force of gravity, resulting in a net of zero work.
A bodybuilder is in the midst of a an intense training session. He is currently bench pressing a bar with a mass of 
. If he does six reps of this mass and his arms are 
long, how much work has been done on the bar between the time the bar was removed from its rack and placed back on the rack?
A bodybuilder is in the midst of a an intense training session. He is currently bench pressing a bar with a mass of
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The most important part of this question is noticing that it asks how much work has been done on the bar, not how much work the bodybuilder has exerted. Therefore we can use the work energy theorem:
Since the bar is initially at rest and returns to rest, the net work on the bar is zero. All of the energy exerted by the bodybuilder is counteracted by gravity.
Think about the system practically. Comparing the initial and final states, the bar is in the exact same position.
The most important part of this question is noticing that it asks how much work has been done on the bar, not how much work the bodybuilder has exerted. Therefore we can use the work energy theorem:
Since the bar is initially at rest and returns to rest, the net work on the bar is zero. All of the energy exerted by the bodybuilder is counteracted by gravity.
Think about the system practically. Comparing the initial and final states, the bar is in the exact same position.
A semi-truck carrying a trailer has a total mass of 
. If it is traveling up a slope of 
to the horizontal at a constant rate of 
, how much power is the truck exerting?
A semi-truck carrying a trailer has a total mass of
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Since the truck is traveling at a constant rate, we know that all of the power exerted by the truck is going into a gain in potential energy. The power exerted will be a function of the change in potential energy over time. Therefore, we can write the following formula:
is a vertical height, so we need to write that as a function of distance traveled up the slope:
We can substitute velocity into this equation:
We have values for all of these variables, allowing us to solve:
Since the truck is traveling at a constant rate, we know that all of the power exerted by the truck is going into a gain in potential energy. The power exerted will be a function of the change in potential energy over time. Therefore, we can write the following formula:
We can substitute velocity into this equation:
We have values for all of these variables, allowing us to solve:
The slingshot is a popular attraction at many amusement parks. During the ride, riders sit in a ball and are propelled vertically by a massive slingshot. A certain sling shot uses 25 springs, each with a constant of 
. If the springs are stretched 
and the ball reaches a maximum height of 
when no riders are in it, what is the total mass of the ball?
The slingshot is a popular attraction at many amusement parks. During the ride, riders sit in a ball and are propelled vertically by a massive slingshot. A certain sling shot uses 25 springs, each with a constant of
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We can use the expression for conservation of energy to solve this problem:
Given the problem statement, we know that the ball has no velocity at the initial and final states, so we can remove kinetic energy from the equation:
The initial potential energy is stored in the springs and the final potential energy is gravitational, so we can write:
We multiply by 25 because there are 25 springs.
Rearranging for mass, we get:
We know all of these values, allowing us to solve:
We can use the expression for conservation of energy to solve this problem:
Given the problem statement, we know that the ball has no velocity at the initial and final states, so we can remove kinetic energy from the equation:
The initial potential energy is stored in the springs and the final potential energy is gravitational, so we can write:
We multiply by 25 because there are 25 springs.
Rearranging for mass, we get:
We know all of these values, allowing us to solve:
An upward force is applied to lift a 
bag a to a height of 
. The bag is lifted at a constant speed. What is the work done on the bag?
An upward force is applied to lift a
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Work is change in energy, so at a height of 
, the bag has more potential energy that when on the ground (zero gravitational potential energy). potential energy is equal to:
Alternatively, 
.
is zero in this case, and 
Work is change in energy, so at a height of
Alternatively,
Fido, a small dog that weighs 
, sees a bird in a tree and climbs straight up, at constant velocity, with an average power of 
. If it takes Fido 
to reach the branch upon which the bird is perched, how high is that branch?
Fido, a small dog that weighs
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We can assume that the dog must carry his entire weight up the tree, and therefore a 
force is exerted. Using the equation
we can use the evidence provided by the problem to solve for distance.
We can assume that the dog must carry his entire weight up the tree, and therefore a
we can use the evidence provided by the problem to solve for distance.
As a joke, Charlie glues C.J's phone to its receiver, which is bolted to her desk. Trying to extricate it, C.J. pulls on the phone with a force of 
for 
. She then pulls on the phone with a force of 
for 
. Unfortunately, all of her exertion is in vain, and neither the phone, nor receiver move at all. How much work did C.J. do on the phone in her 25 total seconds of pulling?
As a joke, Charlie glues C.J's phone to its receiver, which is bolted to her desk. Trying to extricate it, C.J. pulls on the phone with a force of
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Work is a measure of force and displacement 
. Because C.J. did not move the phone at all, no work was done.
Work is a measure of force and displacement
A rocket is in space at location 
when it fires it's thrusters. The thrusters provide a force of 
. The thusters are turned off at location 
.
What is the work done on the rocket by the thrusters?
A rocket is in space at location
What is the work done on the rocket by the thrusters?
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Because all of the force was in the Y direction, all of the work will come from the change in the Y coordinate.
We will use the definition of work
First we need to find the distance traveled.
Then we can plug that into our work equation.
Because all of the force was in the Y direction, all of the work will come from the change in the Y coordinate.
We will use the definition of work
First we need to find the distance traveled.
Then we can plug that into our work equation.
You are pulling a dolly with special frictionless wheels on a level surface. There is a rope from your hand to the dolly which you are using to pull it with. The rope makes a 
angle with the ground. You are applying 
of force.
How much work is done on the dolly over a distance of 
?
You are pulling a dolly with special frictionless wheels on a level surface. There is a rope from your hand to the dolly which you are using to pull it with. The rope makes a
How much work is done on the dolly over a distance of
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The definition of work is
Where
is Force
is displacement
and 
is the angle between the displacement and the force.
The angle will be 
because the dolly is traveling horizontally and the rope which is applying the force is 
above the horizontal.
We plug in our values
The definition of work is
Where
and
The angle will be
We plug in our values
Determine the work done by nonconservative forces if an object with mass 10kg is shot up in the air at 
returns to the same height with speed 
.
Determine the work done by nonconservative forces if an object with mass 10kg is shot up in the air at
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Since this question refers to work done by nonconservative forces, we know that:
Here, 
is the change in potential energy, and 
is a change in kinetic energy. 
is 
because the object returns to the same height as when it was launched. 
however has changed because the object's velocity has changed. Recall that the formula for the change in kinetic energy is given by:
Here 
is the mass of the object, 
is the final velocity of the object and 
is the initial velocity of the object.
In our case:
, 
, and 
Since this question refers to work done by nonconservative forces, we know that:
Here,
Here
In our case:
The work done by a centripetal force on an object moving in a circle at constant speed is .
The work done by a centripetal force on an object moving in a circle at constant speed is .
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Recall that work can be defined as:
Here, 
is the magnitude of the force vector, 
is the magnitude of the displacement vector, and 
is the angle between the directions of the force and displacement vectors. In the case of circular motion, the force vector is normal to the circle since it points inward, and the displacement vector is tangent to the circle. This means that the angle between the force vector and displacement is 
. Since 
, work done by the centripetal force on an object moving in a circle is always 
.
Recall that work can be defined as:
Here,
Raul is pushing a broken down car across the flat expanse of the Mojave to his shop.
If his shop is three kilometers away and he pushes with a Herculean force of one thousand newtons in the direction of his shop, how much work will be done on the car?
Raul is pushing a broken down car across the flat expanse of the Mojave to his shop.
If his shop is three kilometers away and he pushes with a Herculean force of one thousand newtons in the direction of his shop, how much work will be done on the car?
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Work is given by the dot product of force and displacement. Since both the force and displacement are in the same direction in this problem, work is simply the product of the two:
Work is given by the dot product of force and displacement. Since both the force and displacement are in the same direction in this problem, work is simply the product of the two:
Juri is tugging her wagon behind her on the way to... wherever her wagon needs to go. The wagon repair shop. She has a trek ahead of her--five kilometers--and she's pulling with a force of 200 newtons. If she's pulling at an angle of 35 degrees to the horizontal, what work will be exerted on the wagon to get to the repair shop?
Juri is tugging her wagon behind her on the way to... wherever her wagon needs to go. The wagon repair shop. She has a trek ahead of her--five kilometers--and she's pulling with a force of 200 newtons. If she's pulling at an angle of 35 degrees to the horizontal, what work will be exerted on the wagon to get to the repair shop?
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Work exerted on an object is equal to the dot product of the force and displacement vectors, or the product of the magnitudes of the vectors and the sin of the angle between them:
The work exerted on the wagon in this problem is thus:
Work exerted on an object is equal to the dot product of the force and displacement vectors, or the product of the magnitudes of the vectors and the sin of the angle between them:
The work exerted on the wagon in this problem is thus:
A model rocket weighing 5kg has a net propulsion force of 50N. Over a small period of time, the rocket speeds up (with constant acceleration) from an initial velocity of 
to a final velocity of 
. Let us assume that the loss of mass due to fuel consumption is negligible and that the net force is along the direction of motion. How much net work was done on the rocket?
A model rocket weighing 5kg has a net propulsion force of 50N. Over a small period of time, the rocket speeds up (with constant acceleration) from an initial velocity of
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The net work done is equal to the change in kinetic energy. So we must find the kinetic energy at both the initial and final velocities and subtract.
The net work done is equal to the change in kinetic energy. So we must find the kinetic energy at both the initial and final velocities and subtract.
A 50kg man pushes against a wall with a force of 100N for 10 seconds. How much work does the man accomplish?
A 50kg man pushes against a wall with a force of 100N for 10 seconds. How much work does the man accomplish?
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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.
The answer is
Here,
A large block is sitting on the floor. The block is 
. You pull the block with a rope, applying 
of force at a 25 degree angle with respect to the horizon. How much work did you do on the block if you moved it 
? The block moves at a constant speed while you are pulling it.
A large block is sitting on the floor. The block is
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First let's draw a free body diagram of the block with all the forces acting on it.
The block has a weight force which is 
. The normal force is equal and opposite (otherwise the block would be accelerating into the ground or into the air. The applied force is at an angle, and in order to find the work done we need to find the applied force that is pulling the block across the floor. This is:
The work done is just the product of the applied force and the distance the block slides,
Work is an energy, and the units of 
are equal to Joules.
First let's draw a free body diagram of the block with all the forces acting on it.
The block has a weight force which is
The work done is just the product of the applied force and the distance the block slides,
Work is an energy, and the units of
Consider a constant force, given below, that acts on an object as it moves along the path 
, also given below. Calculate the work 
done on the object. The units of the force and path are 
and 
, respectively.
Consider a constant force, given below, that acts on an object as it moves along the path
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In order to find the work done on the object, we need to take the dot product of the force and the path taken. This is a direct calculation.
In order to find the work done on the object, we need to take the dot product of the force and the path taken. This is a direct calculation.
A car of mass 
is accelerated from 
to 
in 2s.
Determine the work done on the car in this time frame.
A car of mass
Determine the work done on the car in this time frame.
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Use the definition of work:
Plug in known values and solve.
Use the definition of work:
Plug in known values and solve.