Work done by a constant force homework

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Homework Help Overview

The discussion revolves around the calculation of work done when lifting a carton of cat litter. The original poster presents a scenario where a 2.5 kg carton is lifted to a height of 0.75 m, using the equation for work related to potential energy.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of work done, questioning the discrepancy between their computed value and the answer provided in the textbook. They explore the implications of the lifting method and whether the carton itself or the contents are considered in the work calculation.

Discussion Status

The discussion is ongoing, with participants expressing confusion over the textbook answer of zero work done. Some have suggested that the lifting method may affect the interpretation of work, while others have confirmed the calculations presented by the original poster.

Contextual Notes

There is mention of a potential misunderstanding regarding the definition of work in the context of lifting an object, as well as the need for clarity on the lifting method used. The original poster's calculation contains a minor error in the height value, which has been noted but not resolved.

paxian
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Homework Statement



Jennifer lifts a 2.5 kg carton of cat litter from the floor to a height of 0.75 m.
(a) How much total work is done on the carton during this operation?

Homework Equations


Work=PE=mg.y


The Attempt at a Solution


Work = 2.5kg*9.8m/s^2*0.57m= 18.4J

However the answer in the back of the book is 0? I don't know how they get to that answer? Please help!
 
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I don't know how they determined 0 either.

Perhaps we could be pedantic and say that work was done on the 2.5 kg of kitty litter, and that very little was done on the comparatively-weightless carton itself? :smile:
 
paxian said:

Homework Statement



Jennifer lifts a 2.5 kg carton of cat litter from the floor to a height of 0.75 m.
(a) How much total work is done on the carton during this operation?

Homework Equations


Work=PE=mg.y


The Attempt at a Solution


Work = 2.5kg*9.8m/s^2*0.57m= 18.4J

However the answer in the back of the book is 0? I don't know how they get to that answer? Please help!

I believe your answer is correct [though you typed 0.57 rather than 0.75 in your post, but clearly calculated correctly. Make sure you looked up the correct answer.
 
PeterO said:
I believe your answer is correct [though you typed 0.57 rather than 0.75 in your post, but clearly calculated correctly. Make sure you looked up the correct answer.

I looked at the right answer... there is part b to this question and that part is consistent with my answer from the back of the book!
 
On a separate aspect, the title you have used here is not a good choice. The box of litter requires a "constant force" to support it only while it is moving with constant acceleration (this includes sitting on the floor, which is an acceleration of zero). We aren't told how the box was lifted; it's very unlikely it was lifted with a constant force. It doesn't matter how fast or slow or erratically it was lifted, it still gains the same P.E. so the same amount of work is done.
 

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