Step By Step answers please

Question

A person pushes a 50 kg crate 40 meters to the right with a horizontal force of 300 N. If the coefficient of static

friction between the crate and the floor is 0.21. (a) How much work is done on the crate by friction? (b) How

much work is done on the crate by gravity? (c) How much work is done on the crate by the person? (d) What is

the total work done on the crate?

Answer 1

Work, W, equals force times distance cosine of the angle between the force and the distance.

Friction equals µF_n , in this case F_n = mg because floor is horizontal, so work done by friction W_f

W_f = µmg x d cos 180° = [(0.21)(50)(9.8) x 40 (-1) = -4,116 J

work done by applied force, F_a

W_Fa = 300 x 40 x (1) = 12,000 J

Work done by gravity, W_g = 0 since cos 90° = 0

Total work, W_T = W_Fa + W_f = 12,000 + (-4,116) = 7,884 J

Answer 2

a. The work done against friction is F_f x D = μ x N x D = 0.21 x 50 kg x 9.8 m/s² x 40 m = 4116 Joules or 4100 J to 2 sig figs.

b. No work is done on the crate by gravity as gravitation force is downward and the crate doesn't move down.

c. The person does 300N x 40 m = 12000 Joules (sum of work on crate and work of friction)

d. Work done on the crate is 12000 J - 4116 J = 7884 J or 7900 J to 2 sig figs.