Answer
(a) $v = \sqrt{2gh - 2g~\mu_k~\sqrt{L^2-h^2}}$
(b) $v = 9.95~m/s$
Work Step by Step
(a) Let $\theta$ be the angle the slope makes with the horizontal. Then $sin(\theta) = \frac{h}{L}$ and $cos(\theta) = \frac{\sqrt{L^2-h^2}}{L}$
We can use a force equation to find an expression for the acceleration as the object slides down the slope.
$\sum F = ma$
$mg~sin(\theta) - F_f = ma$
$mg~sin(\theta) - mg~cos(\theta)~\mu_k = ma$
$a = g~sin(\theta) - g~cos(\theta)~\mu_k$
We can use the expression for the acceleration to find the speed at the bottom.
$v^2 = v_0^2+2aL = 0 + 2aL$
$v = \sqrt{2aL}$
$v = \sqrt{(2L)(g~sin(\theta) - g~cos(\theta)~\mu_k)}$
$v = \sqrt{(2L)(g~\frac{h}{L} - g~\mu_k~\frac{\sqrt{L^2-h^2}}{L})}$
$v = \sqrt{2gh - 2g~\mu_k~\sqrt{L^2-h^2}}$
(b) We can use the expression in part (a) to solve this question.
$v = \sqrt{2gh - 2g~\mu_k~\sqrt{L^2-h^2}}$
$v = \sqrt{(2)(9.80~m/s^2)(12~m) - (2)(9.80~m/s^2)(0.07)~\sqrt{(100~m)^2-(12~m)^2}}$
$v = 9.95~m/s$
