Question

A force F at an angle θ above the horizontal is used to pull a heavy suitcase of weight mg a distance d along a level floor at constant velocity. The coefficient of friction between the floor and the suitcase is µ. The work done by the frictional force is: 

(A) –Fd cos θ

(B) mgh – Fd cos θ

(C) – µ Fd cos θ

(D) -µmgd

(E) -µ mgd cos θ

Answer 1

Correct option (A) –Fd cos θ

Constant velocity → F_net = 0, f_k = Fx = Fcos θ W_fk = – f_kd = – Fcosθd