Answer
a) $\Delta x = \frac{v_0^2}{2(gsin \theta-mu_kgcos\theta)}$
b) $0=\frac{-v_0^2(cos\theta+\mu sin\theta)}{2g(sin\theta - \mu cos\theta)}$
Work Step by Step
a) Using conservation of energy, we find:
$\frac{1}{2}mv_0^2 = mgh-\mu_kmgcos\theta\Delta x$
$\frac{1}{2}v_0^2 = gh-\mu_kgcos\theta\Delta x$
$\frac{1}{2}v_0^2 = g(\Delta x sin \theta)-\mu_kgcos\theta\Delta x$
$\Delta x = \frac{v_0^2}{2(gsin \theta-mu_kgcos\theta)}$
b) To find the minimum angle, we take the derivative and set it equal to zero:
$0=\frac{-v_0^2(cos\theta+\mu sin\theta)}{2g(sin\theta - \mu cos\theta)}$
Here, theta is the minimum angle.
