Question

I've got an energy and work problem. The premise of the problem is:

Answer 1

Refer to the diagram shown below.

μ =  the coefficient of dynamic friction between the crate and the ramp.

1. The applied force of F acts over a distance, d.
    The work done is F*d.

2. The component of the weight of the crate acting down the ramp is
    mg sin(30) = 0.5mg. 
    The work done by this force is 0.5mgd.

3. The normal force is N = mgcos(30) = 0.866mg.
     This force is perpendicular to the ramp, therefore the work done is zero.

4. The frictional force is μN = μmgcos(30) = 0.866μmg.
    The work done by the frictional force is 0.866μmgd.

5. The total force acting on the crate up the ramp is
     F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ) 

6. The work done on the crate by the total force is
    d*(F - 0.5mg - 0.866μmg)