# Pulling a block up an incline: work!

1. Sep 24, 2007

What is the total work W done on the block by the applied force F as the block moves a distance L up the incline?

http://session.masteringphysics.com/problemAsset/1006802/15/2550_a.jpg

isn't it just FLcos(theta) minus the work of friction which is mew(mg)Lcos(theta)??

what am i missing here????

2. Sep 24, 2007

### mgb_phys

Work is just force * distance.
The friction opposing F is simply friction_coef * normal force

3. Sep 24, 2007

first of all the answer must be expressed in terms of F, L, theta, mew, m, and g

second, well... you're not taking into account the angle...

4. Sep 24, 2007

### mgb_phys

Sorry the diagram isn't very clear, I thought the force was acting in the direciton of the red arrow - if it actually acting horizontally as shown by the tiny arrow above the F, then the FLcos(theta) is correct.

5. Sep 24, 2007

But FLcos(theta) is not the answer. shouldn't it be FLcos(theta) minus something like force of friction??

6. Sep 24, 2007

### mgb_phys

Normal force is m g cos(theta), so friction along the slope is coeff_friction * m g cos(theta).
The force up the slope is F cos(theta) Assuming that the little arrow means F is horizontal.
So the work done by the force is ( F cos(theta) - u m g cos(theta) ) times distance, or W = L ( F - u m g ) cos(theta) unless I made a mistake !

7. Sep 24, 2007

that's the answer i originally came up with isn't it (look at first post). but i know for a fact that the answer is wrong.... i'm so confused. anyone know how to solve this thing??

8. Sep 25, 2007