How to calculate work from force applied at downward angle and friction

[Chris Mexas]

Homework Statement

If you exert a force of 175 N at a downward angle of 35 degrees on a box with a total mass of 125kg for a distance of 15m, and the box and the floor have a u_k of 0.25, what is the work done?

Homework Equations

F= (m)(a)
F_k= (u_k) (Fn)
W = (F)(d)

The Attempt at a Solution

1) First I drew a free body diagram: the first horizontal vector will be in the positive direction and will be F=(175)(cos 35) = 143.35 N.
2) I drew a vertical vector in the negative (down) direction, representing the weight of the box F=(125)(9.81) = 1,226.25 N.
3) Since the box is being pushed at an angle, I drew the vertical component of the force, which is also negative (down) F=(175)(sin 35) = 100.38 N.
4) The normal force Fn should be equal to (and in opposite direction to) the two vertical forces calculated in steps 2 and 3. Fn=(1,226.25 + 100.38) = 1326.63 N.
5) The Force due to friction would then be shown as a vector in the negative direction and would be Fk = u_k (FN) = (0.25)(1326.63) = 331.65 N.

Since the force caused by friction is larger than the horizontal force applied to the box, (assuming I did everything right), does this mean that the box is not moving?
The net force would be F=-331.65 + 143.35 = -188.30N

If I apply this to the equation for work I would get W = (-188.30)(15) = -2,824.5 Joules
Is this correct?
Thanks for your help!
Chris

 

[Chris Mexas]

fizzsucks,

The only thing that could be wrong with steps 1 and 3 is my interpretation of downward angle. I'm assuming that a downward angle of 35 degrees means that the force is applied at an angle of 35 degrees with respect to the horizontal axis. In other words, an angle of 0 degrees would be horizontal, an angle of 90 degrees would be vertical. I could try the problem using the same formulas but applying 55 degrees (90-35) instead of applying 35. Is this what you mean?