A 123-kg crate is being pushed across a horizontal floor by a force P that makes an angle of 30.0 ° below the horizontal. The coefficient of kinetic friction is 0.270. What should be the magnitude of P, so that the net work done by it and the kinetic frictional force is zero?
Fk = k*Fn
The Attempt at a Solution
I drew a free body diagram.
| crate ^ normal force
v <---- frictional force (coefficient = .27)
The normal force pushing on the crate is 1205.4 N. Multiplying this by the coefficient of kinetic friction gives me 325.46 N for the magnitude of Fk. This is in opposition to the crate's movement, so the horizontal force pushing back is also 325.46 (making the opposing force negative, really). Plugging this into Pcos@ gives a P value of 375.8 N. However, this is incorrect.
To make the net force zero (and therefore the net work zero), they should be opposites of each other, right? So where is the hole in my logic?