A force of 75.0 N in the horizontal direction is required to set a 37.3-kg box, initially at rest on a flat surface, in motion. After the box begins to move, a horizontal force of 51.3 N is required to maintain a constant speed.

Question

A force of 75.0 N in the horizontal direction is required to set a 37.3-kg box, initially at rest on a flat surface, in motion. After the box begins to move, a horizontal force of 51.3 N is required to maintain a constant speed.
(a) What is the coefficient of static friction between the box and the surface?

(b) What is the coefficient of kinetic friction between the box and the surface?

Scott · Answer

(a) 75.0 / weight ... 75.0 / (37.3 * g) (b) 51.3 / weight ... 51.3 / (37.3 * g)

Henry · Answer

M*g = 37.3 * 9.8 = 365.5 N. = Wt. of box = Normal(Fn).

a. Fs = u*Fn = 365.5u. = Force of static friction.
Fap-Fs = M*a.
75 - 365.5u = M*0.
u = 0.205. = coefficient of static friction.

b. Fap-Fk = M*a.
51.3 - 365.5u = M*0,
u = 0.140. = coefficient of kinetic friction.

Note. Since velocity is constant, acceleration is zero.