Answer
The friction force is 104 N.
The coefficient of kinetic friction $\mu_k$ is 0.48
Work Step by Step
$\sum ~F = ma$
$mg ~sin(\theta) - F_f = ma$
$F_f = mg ~sin(\theta) - ma$
$F_f = (25.0 ~kg)(9.80 ~m/s^2) ~sin(27^{\circ}) - (25.0 ~kg)(0.30 ~m/s^2) = 104 ~N$
The friction force is 104 N. We can use the friction force to find the coefficient of kinetic friction $\mu_k$:
$mg ~cos(\theta)\cdot \mu_k = F_f$
$\mu_k = \frac{F_f}{mg ~cos(\theta)} = \frac{104 ~N}{(25.0 ~kg)(9.80 ~m/s^2) \cdot ~cos(27^{\circ})} = 0.48$
The coefficient of kinetic friction $\mu_k$ is 0.48
