Consider the same 10.0 kg box, still at rest on a level surface with coefficients of friction _s = 0.35 and _k = 0.25. A force of 40.0N is applied to the box at an angle of 35.0 below the horizontal.

(a) Draw a free body diagram of the box and calculate all the forces acting on it in component form. Hint: what will the normal force be, given that the box should not accelerate in the vertical direction?

(b) Explain how you can tell that the applied force will not cause the box to move.

(c) Now imagine that some oil is poured on the floor to reduce the friction, and the box is made to move. The amount of oil is just enough to allow the applied force above to keep the box moving at constant velocity. What is the new coefficient of kinetic friction?

(d) For a displacement of 2.00m, what is the work done by the force of friction?

(e) What is the work done by the applied force? Hint: to check your answer, remember that the velocity is constant and think about what the net work should be.