Pulling a block up an incline: work

[jaded18]

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?

 

[mgb_phys]

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

 

[jaded18]

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...

 

[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.

 

[jaded18]

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

 

[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 !

 

[jaded18]

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??

 

[jaded18]

oh. haha. i get it. use sin instead of cos for the first part with the force of gravity. thanks for trying, mgb_phys!

 

[mckicks88]

so what is the correct answer?