# Homework Help: Work done by frition on a crate!

1. Sep 20, 2009

### cbeeson23

A 70 kg crate starting at rest slides down a rough ramp 4.0 meters long, inclined at 30 degrees above horizontal. At the bottom, it is moving at 5.0 m/s. what was the work done by friction on the crate???

A. -500 Joules
B. 500 Joules
C. 460 Joules
D. -460 Joules
E. 0.0 Joules

I got this answer wrong and i was wondering why... so here is my work. i got the height to be 2meters by using sin 30

Mgh - Wf = 1/2MV^2

70 * 9.81 * 2 - wf = .5 * 70 * 5.0^2
875 - wf = 1373.4
wf =1373.4-875
wf = 500 joules

so i got positive 500 joules. can any body tell me what the answer truly is and what i did wrong?? thank you alot!!

2. Sep 20, 2009

### tiny-tim

Welcome to PF!

Hi cbeeson23! Welcome to PF!

(try using the X2 tag just above the Reply box )
Work done = force "dot" displacement …

the friction is up, but the displacement is down.

3. Sep 20, 2009

### cbeeson23

so the answer then would be -500 joules?

should my equation have been mgh = 1/2mv - work

4. Sep 20, 2009

### cbeeson23

Or are you saying that the work done is actually 500 joules, but since friction is what goes up the ramp the work done by friction is actually - 500 joules?

5. Sep 20, 2009

### cbeeson23

can any1 explain this answer to me a little better

6. Sep 20, 2009

### rl.bhat

Frictional force acts in the opposite direction of the displacement of the crate.
When the work is done on a body, its kinetic energy increases.
When the work is done by a body, its kinetic energy decreases.
When there is no friction, mgh = 1/2*mv^2
Here mgh > 1/2*mv^2
So due to friction kinetic energy of the crate decreases.
Hence work done by the friction = final energy - initial energy.

7. Sep 21, 2009

### tiny-tim

Hi cbeeson23!

(just got up :zzz: … )

If you're confused, then don't treat potential energy as part of the energy, treat it as part of the work done (PE is just another name for minus work done in a conservative field, such as gravity).

The formula to remember is ∆(kinetic energy) = total work done

which in this case is mgh - Fd