# Quick Work & Energy Problem

1. Jun 15, 2014

### ThatDude

1. The problem statement, all variables and given/known data

A person is pushing a box on a floor with a 250 N force. The box is moving at a constant speed over 12.75 m.

(a) What quantity of work does the person do on the box?
(b) What quantity of word does friction do?
(c) What is the total work done?

2. The attempt at a solution

(a) W = F * d * cosx = (250 N)(12.75 m)(cos0) = 3187.5 Joules

(b) W = F * d * cosx = (250 N)(12.75 m)(cos180) = -3187.5 Joules

(c) Total work = Wperson + Wfriction = 3187.5 J + (-3187.5 J) = 0

2. Jun 15, 2014

Do you have a question?
Your working seems correct.(I don't have my calculator at the moment)

3. Jun 15, 2014

### ThatDude

I was wondering if my process for (c) was correct.

4. Jun 15, 2014

### Staff: Mentor

Probably not correct, but now I'm unsure of whether there is a strict convention to adhere to on the topic of friction. Were I to answer this as a test, I would have given the one identical answer to all 3 parts.

I wouldn't like to say zero total work has been done, because I can see that heat has been added to the system as a result of the friction. (So I'm tipping that maybe we say the work done by friction is zero.)

We'd best wait on to see what others say.

Last edited: Jun 15, 2014
5. Jun 15, 2014

### rude man

(b) is a negative number.
(c) = (a) + (b)

6. Jun 16, 2014

### Staff: Mentor

I can see why work done by friction is considered negative. W=F.s, so when motion is in a direction opposite to that of F then one of the terms will have a negative sign. For the block, though, where movement is in the same direction as F then their product will be positive, and positive work is done. Thanks rude man.

7. Jun 16, 2014

That means all the answers in the OP are correct

8. Jun 16, 2014

### rude man

In any case, yes, they are now correct if he did the math correctly.

Last edited: Jun 16, 2014
9. Jun 17, 2014

### Staff: Mentor

The confusion was mine. OP was correct from the start.