Answer
The work done by mg is -240 J.
The work done by $F_p$ is 240 J.
The work done by $F_N$ is 0.
Work Step by Step
Let's find the work done by $mg$ and let's call this $W_G$.
$W_G = mg ~sin(\theta)~d ~cos(180^{\circ})$
$W_G = (16 ~kg)(9.80 ~m/s^2) ~sin(12^{\circ})(7.5 ~m)(-1)$
$W_G = -240 ~J$
Since the cart is moving at a constant speed, the acceleration is zero. Therefore, the component of $F_p$ directed up the ramp must be of equal magnitude to the component of the weight directed down the ramp. The work done by $F_p$ must be equal and opposite to the work done by $mg$.
$W_F = 240 ~J$
The normal force is at a $90^{\circ}$ angle to the direction of motion. Therefore, the normal force does no work on the cart.
