# Work in an elevator

## Homework Statement

Near the surface of the Earth, a block of cheese with mass m lies on the foor of
a elevator cab with mass M = 1,000m (i.e., the mass of the cab is 1,000 times
the mass of the cheese) that is being pulled upward by a cable. If through the
distance d = 2m, the normal force on the block from the floor has constant
magnitude FN = 3N, how much work is done on the cab by the force from the
cable?

W = fd
F = ma

## The Attempt at a Solution

The distance moved is two meters, so the only unknown necessary to solve is the force from the cable on the cab.

If you consider the entire system of cheese and elevator,
Fc - mg - 1000mg = 1001ma

Just the cheese,
3 - mg = ma
a = 3/m - g

Combine the two,
Fc = 1001mg + 1001m(3/m - g)
Fc = 1001mg + 3003 - 1001mg
Fc = 3003 N

W = (3003N)(2m) = 6,006 J

However, the answer is apparently 6,001 J. Can anyone explain why to me? Or is this a misprint on the answer sheet...

collinsmark
Homework Helper
Gold Member
W = (3003N)(2m) = 6,006 J

However, the answer is apparently 6,001 J. Can anyone explain why to me? Or is this a misprint on the answer sheet...
According to my calculations, it must be a mistake on the answer sheet. 'That or something was left out of the problem statement.

I came up with your value, 6006 J for the total work done on the cab+cheese by the cable. I'm assuming that distance d = 2 meters (instead of some sort of weird dependency on mass, m), FN = 3 Newtons, the cable has negligible weight/mass, and the 6006 J of work is the work done on the cab+cheese.

On the other hand, the problem statement says, "how much work is done on the cab by the force from the cable," and doesn't mention anything about the work done on the cheese (by the normal force from the cab). It's sort of ambiguous, but maybe you're supposed to subtract off the work done on the cheese from the total. Personally I wouldn't interpret it that way. The way I see it, 6006 J of work is done on the cab, but the cab does 6 J of work on the cheese, which leaves a net of 6000 J done on the cab by itself (ignoring the work done on the cheese), even though the work done by the cable's force is still 6006 J. But even if you're supposed to give the net amount of work done on the cab, by itself, still doesn't quite give an answer of 6001 J either.

So I'm guessing that there's either something left out of the problem statement, or there is a mistake in the answer sheet.