kinetic friction problem?

Question

A stockroom worker pushes a box with mass 11.0 kg on a horizontal surface with a constant speed of 3.60 m/s. The coefficient of kinetic friction between the box and the surface is 0.240

What horizontal force must be applied by the worker to maintain the motion? (i already calculated this answer to be 25.9N)

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

Answer 1

When the force is removed, the box has kinetic energy of 0.5*m*v^2. The work done by friction is F*d, where F is the frictional force and d is the distance. F = µ*m*g. The box stops when the frictional work equals the initial kinetic energy, or 0.5*m*v^2 = µ*m*g*d. Then

d = 0.5*v^2 / µ*g = 2.753m