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) -µ Fd cos θ
(C) -µmgd
(D) -µ mgd cos θ

Answer 1

Answer:

option A

Explanation:

given,

Force = F

angle = θ

weight on suitcase = mg

distance = d

constant velocity so, acceleration a = 0

coefficient of friction = µ

Work done = ?

Work done is equal to force into displacement.

Friction act opposite to the force acting so, work done by frictional force will be negative.

frictional force will act into horizontal direction opposite to force.

here displacement is equal to d

now,

W = -F d cos θ

Hence,the correct answer is option A