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

A crate with a mass of m is on a ramp that is inclined at an angle of 30° from the horizontal. A force with a magnitude of F directed parallel to the ramp is used to pull the crate with a constant speed up the ramp a distance of d.

There are a set of subquestions and I've answered them as best as I can, but I don't think I'm right. Could someone give me the answers and help me understand why I'm wrong?

1. What is the work done on the crate by the applied force F?
--I said cos30 * F * displacement

2. 2. What is the work done on the crate by the gravitational force exerted on the crate by Earth?
--I said M*9.8*tan30

3. What is the work done on the crate by the normal force, with a magnitude of Fn, exerted on the crate by the ramp? (Hint: recall that the normal force is perpendicular to the surface of the ramp.)
--I said zero

4. What is the work done on the crate by the frictional force Fk?
--I don't understand this one

5. What is the total force acting on the crate?
--I said F, as in the pre-defined force 'F'

6. What is the work done on the crate by the total force?
--I don't understand how this is different from the first one

Respuesta :

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)

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