Answer
(a) The coefficient of static friction $\mu_s$ between the box and the floor is 0.60.
(b) The coefficient of kinetic friction $\mu_k$ between the box and the floor is 0.53.
Work Step by Step
(a)
$F_f = 35.0 N$
$mg ~\mu_s = 35.0 ~N$
$\mu_s = \frac{35.0 ~N}{(6.0 ~kg)(9.80 ~m/s^2)} = 0.60$
The coefficient of static friction $\mu_s$ between the box and the floor is 0.60.
(b)
$\sum F = ma$
$F - F_f = ma$
$mg ~\mu_k = F -ma$
$\mu_k = \frac{F-ma}{mg} = \frac{35.0 ~N - (6.0 ~kg)(0.60 ~m/s^2)}{(6.0 ~kg)(9.80 ~m/s^2)} = 0.53$
The coefficient of kinetic friction $\mu_k$ between the box and the floor is 0.53.
