Answer
The speed at the top of the ramp is 3.17 m/s.
Work Step by Step
The kinetic energy at the top will be equal to the kinetic energy at the bottom plus the work done by the force and gravity.
$\frac{1}{2}mv_2^2 = \frac{1}{2}mv_1^2 + F~cos(\theta)~d - mg~sin(\theta)~d$
$v_2^2 = \frac{mv_1^2 + 2F~cos(\theta)~d - 2mg~sin(\theta)~d}{m}$
$v_2 = \sqrt{\frac{mv_1^2 + 2F~cos(\theta)~d - 2mg~sin(\theta)~d}{m}}$
$v_2 = \sqrt{\frac{(85.0~kg)(2.00~m/s)^2 + (2)(600~N)~cos(30.0^{\circ})(2.50~m) - (2)(85.0~kg)(9.80~m/s^2)~sin(30.0^{\circ})(2.50~m)}{85.0~kg}}$
$v_2 = 3.17~m/s$
The speed at the top of the ramp is 3.17 m/s.
