#### Answer

(a) $v = \sqrt{\frac{2~F~h~cot(\theta) - 2mgh}{m}}$
(b) v = 4.0 m/s

#### Work Step by Step

(a) We can find an expression for the distance $d$ that the crate moves along the slope;
$\frac{h}{d} = sin(\theta)$
$d = \frac{h}{sin(\theta)}$
The work done by the force will be equal to the sum of the potential energy and the kinetic energy. Therefore;
$PE+KE = W$
$mgh + \frac{1}{2}mv^2 = F~d~cos(\theta)$
$\frac{1}{2}mv^2 = F~(\frac{h}{sin(\theta)})~cos(\theta) - mgh$
$v^2 = \frac{2~F~h~cot(\theta) - 2mgh}{m}$
$v = \sqrt{\frac{2~F~h~cot(\theta) - 2mgh}{m}}$
(b) We can use the expression in part (a) to find the speed at the top of the slope:
$v = \sqrt{\frac{2~F~h~cot(\theta) - 2mgh}{m}}$
$v = \sqrt{\frac{(2)(25~N)(2.0~m)~cot(20^{\circ}) - (2)(5.0~kg)(9.80~m/s^2)(2.0~m)}{5.0~kg}}$
$v = 4.0~m/s$