#### Answer

(a) The speed at the bottom of the ramp is $\sqrt{2gd~sin(\alpha)}$
(b) The speed at the bottom of the ramp is $\sqrt{2gd~sin(\alpha)}$

#### Work Step by Step

(a) $K_2 = U_{grav,1}$
$\frac{1}{2}mv_2^2 = mgh$
$v_2^2 = 2gd~sin(\alpha)$
$v_2 = \sqrt{2gd~sin(\alpha)}$
The speed at the bottom of the ramp is $\sqrt{2gd~sin(\alpha)}$.
(b) $K_2 + U_{grav,2}= K_1+U_{grav,1}$
$\frac{1}{2}mv_2^2 + mg(-h) = 0$
$\frac{1}{2}mv_2^2 = mgh$
$v_2^2 = 2gd~sin(\alpha)$
$v_2 = \sqrt{2gd~sin(\alpha)}$
The speed at the bottom of the ramp is $\sqrt{2gd~sin(\alpha)}$
(c) The normal force didn't enter the solution because the normal force acts at a $90^{\circ}$ angle to the direction of motion, so the normal force does zero work. Also, the ramp is frictionless.