A 53 N force is needed to slide a 50.0kg box across a flate surface at...
A 53 N force is needed to slide a 50.0kg box across a flate surface at a constant velocity. What is the coefficient of kinetic friction between the box and the floor?
0.10, 0.13, 0.09, 0.11
Homework Answers
Since the box is moving with constant velocity...means that the
body is in dynamic equilibrium ..net force acting on box =0 N
kinetic friction force must equal to 53 N but opposite in direction
then kinetic frictional force = 53 N
mu_k*m*g = 53
coefficient of friction is mu_k = 53/(m*g) = 53/(50*9.8) = 0.108
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