Question

A stockroom worker pushes a box with mass 11.9 kg on a horizontal surface with a constant speed of 3.40 m/s . The coefIficient of kinetic friction between the box and the surface is 0.17.

What horizontal force must the worker apply to maintain the motion?

If the force calculated in the previous part is removed, how far does the box slide before coming to rest?

Answer 1

The weight of the box is: W = mg

Since the only other vertical force is the normal force, the normal force is:   N = W = mg

Then, the friction force is:   Ff = u*N = umg

Since the only other horizontal force is the applied force, the applied force is:
F = Ff = umg = (0.17)*(11.9 kg)(9.8 m/s^2) = 19.83 N

The kinetic energy of the box before the force is removed is: K = (1/2)mv^2

The work that friction does to stop the box is: W = -Ff*d = -umgd

Since the final kinetic energy of the box is 0:
K + W = 0
(1/2)mv^2 + -umgd = 0
(1/2)mv^2 = umgd
d = v^2/(2ug) = (3.4 m/s)^2/[2*(0.17)(9.8 m/s^2)] = 3.5 m