Answer
The block's speed is $~~2.2~m/s$
Work Step by Step
In part (b), we found that the block travels $~~d = 1.5~m~~$ along the plane.
We can find the work done by the frictional force as the block goes up its the highest point:
$W = -mg~cos~\theta~\mu_k~d$
$W = -(5.0~kg)(9.8~m/s^2)(0.40)(1.5~m)~(cos~30^{\circ})$
$W = -25.5~J$
Since the block goes up and then returns to the bottom, the total work done by the frictional force is $~~(2)(-25.46~J)~~$ which is $~~-50.9~J$
We can use work and energy to find the speed at the bottom:
$K_2 = K_1+W$
$\frac{1}{2}mv_2^2 = \frac{1}{2}mv_1^2+W$
$v_2^2 = v_1^2+\frac{2W}{m}$
$v_2 = \sqrt{v_1^2+\frac{2W}{m}}$
$v_2 = \sqrt{(5.0~m/s)^2-\frac{(2)(50.9~J)}{5.0~kg}}$
$v_2 = 2.2~m/s$
The block's speed is $~~2.2~m/s$
