Clarification on this specific problem (work done on incline plane)

[elsternj]

Homework Statement

A block of weight sits on a plane inclined at an angle \theta as shown. (Intro 1 figure) The coefficient of kinetic friction between the plane and the block is \mu.

What is the work done by the applied force of magnitude ?
Express your answer in terms of some or all of the following: \mu,w,\theta, L

Homework Equations

The Attempt at a Solution

now the answer to this is W = w(sin(\theta)+\mucos(\theta))L

now I somewhat understand this because the force has to be enough to overcome friction and the weight pulling it down. but what I am concerned with is this:

This would be the answer if the sum of the forces were equal to 0 right?

because if:
\sumF_x=F-wsin\theta-\muwcos\theta=0

when you solve for F you get the answer originally posted and then just multiply it by the length. How is one to know, given the above information, to set that equal to 0 and solve? It doesn't mention anything about moving at a constant speed. Just trying to get some clarification because I always set the sum of my forces equal to ma and I had trouble answering this question because the answer had to only include the variables \mu,w,\theta, L ... so the only way I could see to solve for F and keep only those variables would be if acceleration happened to be 0.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[tiny-tim]

hi elsternj!

A block of weight sits on a plane inclined at an angle \theta as shown. (Intro 1 figure) The coefficient of kinetic friction between the plane and the block is \mu.

What is the work done by the applied force of magnitude ?
Express your answer in terms of some or all of the following: \mu,w,\theta, L
…
It doesn't mention anything about moving at a constant speed.
L ... so the only way I could see to solve for F and keep only those variables would be if acceleration happened to be 0.

yes, you're right, it's a rubbish question