Answer
(a) W = mgh
(b) $v = \sqrt{2gh}$
(c) v = 17.1 m/s
Work Step by Step
(a) The component of gravity parallel to the plane is $mg~sin(\alpha)$. The work done is $mg~sin(\alpha)~d$, where $d$ is the distance along the plane. Note that $d = \frac{h}{sin(\alpha)}$. Then:
$W = mg~sin(\alpha)~d$
$W = mg~sin(\alpha)\frac{h}{sin(\alpha)}$
$W = mgh$
The component of gravity perpendicular to the plane is $mg~cos(\alpha)$. Since it is perpendicular to the direction of motion, the work done is zero.
The total work done by gravity is $W = mgh$, which is the same as if the mass had fallen straight down from a height of $h$.
(b) $KE_1 + W = KE_2$
$0 + mgh = \frac{1}{2}mv^2$
$v = \sqrt{2gh}$
Since the work done by gravity does not depend on the angle $\alpha$, the amount of kinetic energy at the bottom only depends on the height $h$. Thus the speed at the bottom does not depend on the angle $\alpha$.
(c) $v = \sqrt{2gh}$
$v = \sqrt{(2)(9.80~m/s^2)(15.0~m)}$
$v = 17.1~m/s$
