Question

A force of 30.0 N is required to start a 3.8 kg box moving across a horizontal concrete floor
If the 30.0 N force continues, the box accelerates at 0.50 m/s2. What is the coefficient of kinetic friction?

Answer 1

First of all, the static coefficient of friction is
mus = 30/(3.8*9.8)= 0.806

After motion starts, the friction force f is given by:
(30 - f) = M a = 9.8 * 0.5 = 4.9 N
f = 25.1 N
muk = f/(M*g) = 25.1/(9.8*3.8) = 0.675

Answer 2

- (0.5m/s^2)*(3.8kg) = 1.9N
Because Force = (mass)*(acceleration)
- Force of Friction = (30N)-(1.9N) = 28.1N
Because the 30N force is being accelerated therefore the force is does not have to be as strong so the force ultimately decreases.
- Force of Surface = (3.8kg)(9.8m/s^2) = 37.24N
28.1N = (37.24N)*(mu)
mu = 0.755
Because Force of Friction = (Coefficient of Friction)*(Force of Surface)