Lifting an object, kinetic energy, work and forces

In summary: So technically, the work done by gravity and the work done by throwing the object up are both negative, but the net work is zero because there is no change in KE.
  • #1
Karagoz
52
5

Homework Statement


From a physics multiple choice test: "A car is raised to a height of 10 meters. The car is at rest both before and after it has been lifted. The mass of the car is 900 kg. The total work that has been done on the car during the move is"
(The right answer is: zero).

Wnet =ΔKE (Net work = Change in Kinetic energy). The car's kinetic energy remains the same (it's still on rest), so in this equation no work is done on the car.

But how is it that the work done on the car is zero?

Doesn't it contradict with the equation: W = F*s?

A force upwards is applied to the car to raise it up, it requires force against the gravity force to raise the car. And the car is displaced 10 meters upwards with that force.

LiftedCar.png


The force exerted on the car to lift it up is: 900KG*9.81 = 8,829 N.

So according to this equation: W = F*s the work is done on the car, because it's been displaced by a force.

The car is still on rest (no kinetic energy), but the car's potential energy is increased by ca 88 Kj (mgh = 900Kg*9.81*10m). Doesn't it mean that it's been done some work to increase the car's potential energy?

Homework Equations


Wnet =ΔKE (Net work = Change in Kinetic energy). The car's kinetic energy is the remains the same (it's still on rest), so in this equation no work is done on the car.

But according to this equation: W = F*s the work is done on the car, because it's been displaced by a force.

The Attempt at a Solution


I read some answers in Stack Exchange and PhysicsForums, but that just confuses even more:

https://www.physicsforums.com/threads/work-done-to-lift-an-object.828424/

https://www.physicsforums.com/threads/work-done-in-lifting-an-object-against-gravity.624909/

https://www.physicsforums.com/threads/work-done-lifting-an-object-1m-simple-question.477475/
 

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  • #2
Hello @Karagoz,

Welcome to PF! :welcome:

Karagoz said:

Homework Statement


From a physics multiple choice test: "A car is raised to a height of 10 meters. The car is at rest both before and after it has been lifted. The mass of the car is 900 kg. The total work that has been done on the car during the move is"
(The right answer is: zero).

Wnet =ΔKE (Net work = Change in Kinetic energy). The car's kinetic energy remains the same (it's still on rest), so in this equation no work is done on the car.

But how is it that the work done on the car is zero?

Doesn't it contradict with the equation: W = F*s?

A force upwards is applied to the car to raise it up, it requires force against the gravity force to raise the car. And the car is displaced 10 meters upwards with that force.

View attachment 220343

The force exerted on the car to lift it up is: 900KG*9.81 = 8,829 N.

So according to this equation: W = F*s the work is done on the car, because it's been displaced by a force.

The car is still on rest (no kinetic energy), but the car's potential energy is increased by ca 88 Kj (mgh = 900Kg*9.81*10m). Doesn't it mean that it's been done some work to increase the car's potential energy?

Homework Equations


Wnet =ΔKE (Net work = Change in Kinetic energy). The car's kinetic energy is the remains the same (it's still on rest), so in this equation no work is done on the car.

But according to this equation: W = F*s the work is done on the car, because it's been displaced by a force.

The Attempt at a Solution


I read some answers in Stack Exchange and PhysicsForums, but that just confuses even more:

https://www.physicsforums.com/threads/work-done-to-lift-an-object.828424/

https://www.physicsforums.com/threads/work-done-in-lifting-an-object-against-gravity.624909/

https://www.physicsforums.com/threads/work-done-lifting-an-object-1m-simple-question.477475/
This is a somewhat tricky problem due to the wording. Technically, the book is correct about the total work done on the car is 0. It's important to pay careful attention to the wording of "the total work that has been done on the car during the move is ___."

The correct answer would be different if the question was:
"the total work that has been done on the car, by the crane, during the move is ___."

Similarly, yet a different answer than that would be correct for the question:
"the total work that has been done on the car, by gravity, during the move is ___."

You are correct that you can use the formula [itex] W = \vec F \cdot \vec s [/itex] for each force involved in the move.

But how many forces are there for this problem?

Keep in mind the result of the vector dot product can be positive or negative. To find the "total" work done on the car you need to sum the work done by all the forces involved.
 
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  • #3
The work done on an object is its change in KE. The change in PE does not count because that is work done on the car-Earth system.
If you throw an object up, you do work on the object to give it KE, but then gravity does negative work on it to bring it back to rest at the top.
 
  • #4
collinsmark said:
Hello @Karagoz,

Welcome to PF! :welcome:This is a somewhat tricky problem due to the wording. Technically, the book is correct about the total work done on the car is 0. It's important to pay careful attention to the wording of "the total work that has been done on the car during the move is ___."

The correct answer would be different if the question was:
"the total work that has been done on the car, by the crane, during the move is ___."

Similarly, yet a different answer than that would be correct for the question:
"the total work that has been done on the car, by gravity, during the move is ___."

You are correct that you can use the formula [itex] W = \vec F \cdot \vec s [/itex] for each force involved in the move.

But how many forces are there for this problem?

Keep in mind the result of the vector dot product can be positive or negative. To find the "total" work done on the car you need to sum the work done by all the forces involved.

So this means:

Let's call the gravity force on the car (or weight on the car) G, and force from the crane to the car Fc.

1) When the car is lifted up, it needs to be accelerated upwards. Sum of the forces is positive upwards in some seconds.
In the start Fc > G.

2) Then it's lifted upwards with constant velocity. Sum of the forces is zero.
So Fc = G.

3) After 10 meters upwards lifting the car is kept there at rest. There's an acceleration downwards since the car's velocity changes from moving upwards to being at rest.
That means after the car is lift 10 meters upwards sum of the forces is pointing downwards in some seconds to stop car moving even further.
So Fc < G. Fc is 500 N in a short time.

4) The car is kept at rest there 10 meter above the ground. So at there Fc = G again.

This is how the Force and distance graph would look like (Y axis: Sum Forces applied on the car. X axis: meters above the ground).

Force Graph.png


Work is integral of Force with distance as variable. And integral of the graph above is zero (from 0 to 10 meters). This means zero work done on the car (sum is zero).

But there's some areal between the graph and the X axis. Does it mean both the gravity and the crane does work on the car during the lift, but their work done on the car cancels each other out, so the sum becomes zero?
 

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  • #5
Karagoz said:
So this means:

Let's call the gravity force on the car (or weight on the car) G, and force from the crane to the car Fc.

1) When the car is lifted up, it needs to be accelerated upwards. Sum of the forces is positive upwards in some seconds.
In the start Fc > G.

2) Then it's lifted upwards with constant velocity. Sum of the forces is zero.
So Fc = G.

3) After 10 meters upwards lifting the car is kept there at rest. There's an acceleration downwards since the car's velocity changes from moving upwards to being at rest.
That means after the car is lift 10 meters upwards sum of the forces is pointing downwards in some seconds to stop car moving even further.
So Fc < G. Fc is 500 N in a short time.

4) The car is kept at rest there 10 meter above the ground. So at there Fc = G again.

This is how the Force and distance graph would look like (Y axis: Sum Forces applied on the car. X axis: meters above the ground).

View attachment 220383

Work is integral of Force with distance as variable. And integral of the graph above is zero (from 0 to 10 meters). This means zero work done on the car (sum is zero).

But there's some areal between the graph and the X axis. Does it mean both the gravity and the crane does work on the car during the lift, but their work done on the car cancels each other out, so the sum becomes zero?
That's pretty much correct in my opinion.

A note about the work done on the car by the crane: You are right that for at least a brief moment, Fc > g at the beginning and another moment where Fc < g at the end when the car comes to rest at the top. But when calculating the work done by the crane, you can use its average force (or when calculating the force, by using its average acceleration). [Edit: or as you suggest, you can integrate.]

And like @haruspex mentioned, you can use this idea to convince yourself that the net work on the car (same thing as total work once you sum everything together) is equivalent to the car's change in KE.
 
  • #6
collinsmark said:
That's pretty much correct in my opinion.

A note about the work done on the car by the crane: You are right that for at least a brief moment, Fc > g at the beginning and another moment where Fc < g at the end when the car comes to rest at the top. But when calculating the work done by the crane, you can use its average force (or when calculating the force, by using its average acceleration). [Edit: or as you suggest, you can integrate.]

And like @haruspex mentioned, you can use this idea to convince yourself that the net work on the car (same thing as total work once you sum everything together) is equivalent to the car's change in KE.

But isn't it that "the total work done on the car is zero" but "the work done on the car by the care" is not zero (above zero)?
 
  • #7
Karagoz said:
But isn't it that "the total work done on the car is zero" but "the work done on the car by the care" is not zero (above zero)?
The work done by the crane is nonzero.
If we ignore the transient acceleration and deceleration, whether you consider that the crane did zero work on the car but nonzero work against gravity, or that the crane did nonzero work on the car and the car did the same work against gravity, is somewhat arbitrary.
Of course, the crane could in principle pull the car up so fast initially that the car will rise to the required height with no further assistance. In this case clearly the crane does nonzero work on the car, then the car does the same work against gravity.
 
  • #8
collinsmark said:
That's pretty much correct in my opinion.

A note about the work done on the car by the crane: You are right that for at least a brief moment, Fc > g at the beginning and another moment where Fc < g at the end when the car comes to rest at the top. But when calculating the work done by the crane, you can use its average force (or when calculating the force, by using its average acceleration). [Edit: or as you suggest, you can integrate.]

And like @haruspex mentioned, you can use this idea to convince yourself that the net work on the car (same thing as total work once you sum everything together) is equivalent to the car's change in KE.

Know it's an old thread. My intention isn't to keep the thread active forever. But the topic is similar, so instead of new thread I thought to post the question on the same thread.

But there's one thing I wonder:

When the object raised 10 meters high above the ground and kept there, the object now has zero kinetic energy (as it has 0 velocity). But its potential energy is now mg*10. At the ground the object had 0 potential and 0 kinetic energy. 10 meters above the ground the object has Ep = mg*10 (potential energy of mg*10).

Work done by Force = mgh. Work done by Gravity =−mgh.
Net Work Done on the object =mgh−mgh=0 N. But potential energy of the object =mgh.

So the object did get increased its energy (by increasing its potential energy). But what's the source of that energy when the net work done on the object is zero? Where does that object got that "potential energy" from?

How "total work done on the object" can be zero when the object gets increased its energy (by increasing its potential energy)?

According to work-kinetic energy theorem, the net work done on an object is equal to its change in kinetic energy. But why change in potential energy isn't included?
 
Last edited:
  • #9
Karagoz said:
why change in potential energy isn't included?
Because that work does not "reside" in the object. It is embodied in the car-Earth system.
 
  • #10
haruspex said:
Because that work does not "reside" in the object. It is embodied in the car-Earth system.

So an object can't have store potential energy?

But in physics courses, like in this site:
http://www.physicsclassroom.com/class/energy/Lesson-1/Potential-Energy

They wrote:
An object can store energy as the result of its position. For example, the heavy ball of a demolition machine is storing energy when it is held at an elevated position. This stored energy of position is referred to as potential energy. Similarly, a drawn bow is able to store energy as the result of its position. When assuming its usual position (i.e., when not drawn), there is no energy stored in the bow. Yet when its position is altered from its usual equilibrium position, the bow is able to store energy by virtue of its position. This stored energy of position is referred to as potential energy. Potential energy is the stored energy of position possessed by an object.

So it's not correct to say "object A stores potential energy"?

How does this formula:
upload_2018-2-28_21-37-46.png


(When an object is influenced by forces, the sum of all the work of the forces is equal to the change of the kinetic energy of the object)

does fit into this formula?:

upload_2018-2-28_21-38-49.png


(When an object moves in the gravity field, the mechanical energy E is ultimately equal to the mechanical energy E0 at the start plus the work performed by all forces other than the gravity).
 

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  • #11
Karagoz said:
So it's not correct to say "object A stores potential energy"?
Yes, it is not correct.
Re the two formulae, the first is looking at the KE of the object, whereas the second is concerned with the energy of a system of bodies that can move in relation to each other.
 
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  • #12
If you lift an object at constant speed the net force on the object is zero. So the _net_ work done on the _object_ by all the forces must also be zero. Clearly the machine doing the lifting does work but it does this "against gravity" rather than "on the object"...

If you look at the forces on the object you see that the lifting force is in the same direction as the motion so the vector product is positive so we can say the _lifting force_ did positive work on the object. However the force of gravity acts in the opposite direction to the motion so the vector product is negative. We say gravity did negative work on the object. So overall the net work done on the object was zero.

It's one of those horrible questions where the exact wording matters as others have said. It would probably catch me out in an exam.
 

Related to Lifting an object, kinetic energy, work and forces

1. What is the difference between lifting an object and increasing its kinetic energy?

Lifting an object involves moving it against the force of gravity, while increasing its kinetic energy involves increasing its speed or movement. Lifting an object requires work to be done, while increasing its kinetic energy requires a force to act on the object.

2. How does the mass of an object affect the amount of work needed to lift it?

The mass of an object directly affects the amount of work needed to lift it. The greater the mass of an object, the more work is required to lift it against the force of gravity. This is because the force of gravity is directly proportional to an object's mass.

3. What is the relationship between work and kinetic energy?

Work and kinetic energy are directly related. The work done on an object results in an increase in its kinetic energy. This is described by the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.

4. Can an object have kinetic energy even when it is not moving?

Yes, an object can have kinetic energy even when it is not moving. This is known as potential energy and it is the energy an object possesses due to its position or state. For example, a book sitting on a shelf has potential energy because it has the potential to fall and gain kinetic energy when it is dropped.

5. What is the role of forces in lifting an object?

Forces are necessary to lift an object. The force of gravity pulls an object towards the ground, and in order to lift the object, an external force must be applied in the opposite direction. This external force must be equal to or greater than the force of gravity in order to lift the object.

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