Work against gravity confusion

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Discussion Overview

The discussion revolves around the concept of work done against gravity in the context of lifting an object with a tension force. Participants explore the implications of different forces acting on the object, particularly focusing on the relationship between tension, gravitational force, and the resulting work done during the lifting process.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant notes that if the tension is greater than the gravitational force (mg), the object is not only lifted but also accelerated, resulting in additional kinetic energy.
  • Another participant suggests that the question may be poorly worded if it does not specify that the object is moving at constant velocity, which would imply that tension equals mg.
  • A different viewpoint states that work done by tension can be expressed as T*s J, while work done against friction is -F*s J, leading to a discussion about the total work done on the object.
  • One participant emphasizes that "lifting" implies the object is motionless at both the start and end of the displacement, which affects the interpretation of the work done.
  • Another participant reiterates the idea that if the tension equals mg, the total work done on the object would be zero due to the cancellation of work done by tension and gravity, while the work done by tension alone would be mgh J.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the work done in the context of tension and gravitational forces. There is no consensus on whether the original question is poorly worded or if it accurately reflects the physics involved.

Contextual Notes

Participants highlight the importance of assumptions regarding the motion of the object (e.g., constant velocity vs. acceleration) and the definitions of work in different contexts, which remain unresolved.

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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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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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?
 
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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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
 
You would say that force T does T*s J work, while work done by friction is -F*s J.
 
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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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.
 
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.
 

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