Work done by a constant force homework

[paxian]

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!

 

[NascentOxygen]

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?

 

[PeterO]

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.

 

[paxian]

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!

 

[NascentOxygen]

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.