Answer
The normal force is $152.8~N$
The magnitude of the force of kinetic friction exerted on the crate is $61.1~N$ and this force is directed up the ramp at an angle of $30^{\circ}$ to the horizontal.
Work Step by Step
The normal force $F_N$ is the component of the crate's weight that is directed straight into the surface of the ramp.
$F_N = mg~cos~\theta$
$F_N = (18.0~kg)(9.80~m/s^2)~cos~30^{\circ}$
$F_N = 152.8~N$
The normal force is $152.8~N$
Since the crate sliding down the ramp, the force of kinetic friction $F_f$ is exerted on the crate and it is directed up the ramp. We can find the force of kinetic friction:
$F_f = F_N~\mu_k$
$F_f = mg~cos~\theta~\mu_k$
$F_f = (18.0~kg)(9.80~m/s^2)~cos~30^{\circ}~ (0.40)$
$F_f = 61.1~N$
The magnitude of the force of kinetic friction exerted on the crate is $61.1~N$ and this force is directed up the ramp at an angle of $30^{\circ}$ to the horizontal.
