Work against gravity confusion

In summary: Therefore, the question should have specified which work it is referring to. In summary, the first question in the book is not clear as it does not specify which work it is asking for. It can either be the total work done on the object, which would be zero if the tension is equal to mg, or the work done by the tension on the object, which would be mgh J. This lack of clarification can lead to confusion and different interpretations of the answer.
  • #1
Kinhew93
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Hi I've just started learning work, energy etc and the first question in the book is in the form:

An object of mass m is lifted by a tension through a displacement of h. Find the work done on the object.

I know that the answer is simply mgh J, there is something that I don't quite get.

Why can't it be that the tension could be greater than mg but still lifting the object the same distance (in less time) so therefore does more work? I hope that makes sense.

Thanks :)
 
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  • #2
You are correct. If the tension is greater than mg then the object is not only lifted. It is accelerated. It ends with more kinetic energy than it had when it started.

The extra work done in this case manifests as extra kinetic energy.
 
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  • #3
jbriggs444 said:
You are correct. If the tension is greater than mg then the object is not only lifted. It is accelerated. It ends with more kinetic energy than it had when it started.

The extra work done in this case manifests as extra kinetic energy.

Ok so is this just a case of a poorly worded question, as it does not mention that the object is moving at constant velocity? (the answer in the book is mgh)

Would the actual answer be that the tension does at least mgh J of work?
 
  • #4
The "constant velocity" bit would be key. That would allow you to deduce (through Newton's second law) that the tension is equal to mg.

So it does not sound as if the problem was poorly worded.
 
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  • #5
Say there is an object on which a force of T acts for a displacement s and there us a constant frictional force of F.

Would you say that the force T does Ts J of work or that the force T does Ts J of work and Fs J of work against friction? Or is it something else?

Any explanation would be much appreciated :)

Thanks
 
  • #6
You would say that force T does T*s J work, while work done by friction is -F*s J.
 
  • #7
The question said the object was lifted a height h. If I threw a rock so that it passed, say, 10 meters height while still going up to a maximum height of 20 meters, I would NOT say that I "lifted" it 10 m. "Lifted" implies that it was motionless at its original position and motionless at the end. Any acceleration or velocity it had in between is irrelevant.
 
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  • #8
Kinhew93 said:
Say there is an object on which a force of T acts for a displacement s and there us a constant frictional force of F.

Would you say that the force T does Ts J of work or that the force T does Ts J of work and Fs J of work against friction? Or is it something else?

Any explanation would be much appreciated :)

Thanks

For this question, actually the Ts J of work and the Fs J of work is just one same work. It is because the work done by T is used to cancel out the work done by friction. Therefore, the total work done on the object is actually zero.
 
  • #9
Kinhew93 said:
Hi I've just started learning work, energy etc and the first question in the book is in the form:

An object of mass m is lifted by a tension through a displacement of h. Find the work done on the object.

I know that the answer is simply mgh J, there is something that I don't quite get.

Why can't it be that the tension could be greater than mg but still lifting the object the same distance (in less time) so therefore does more work? I hope that makes sense.

Thanks :)

So for this question, I believe that the question is poorly worded. For the sentence "Find the work done on the object", assuming that the tension is same as mg, if it means the TOTAL work done on the object, then the work done will be zero. It's because the work done by both the tension and gravity will cancel each other out. Meanwhile, if it means the work done by the tension on the object, then only the answer will be mgh J.
 

FAQ: Work against gravity confusion

What is work against gravity confusion?

Work against gravity confusion is a term used to describe the difficulty in understanding how work is done when an object is lifted or moved against the force of gravity. It is a common misconception that more work is done when an object is lifted higher, when in fact, the amount of work done is determined by the force applied and the distance the object is moved.

Why is understanding work against gravity confusion important?

Understanding work against gravity confusion is important because it is a fundamental concept in physics and is necessary for understanding more complex concepts, such as potential and kinetic energy. It is also important in practical applications, such as lifting heavy objects or calculating the efficiency of machines.

How is work against gravity confusion different from work on a level surface?

Work against gravity confusion is different from work on a level surface because in the former, the object is being moved against the force of gravity, while in the latter, the object is being moved on a flat surface with no change in elevation. This means that work against gravity requires more energy to be done since the force of gravity must be overcome.

How can work against gravity confusion be calculated?

Work against gravity confusion can be calculated by multiplying the force applied to the object by the distance the object is moved against the force of gravity. This can be represented by the equation W = Fd, where W is the work done, F is the force applied, and d is the distance moved.

What are some real-world examples of work against gravity confusion?

Some real-world examples of work against gravity confusion include lifting weights at the gym, climbing stairs, and pushing a cart uphill. In all of these situations, the object is being moved against the force of gravity, and therefore, work is being done to overcome that force.

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