Answer
a) 20.1 meters per second
b) 30.4 meters per second
Work Step by Step
We know that the energy lost to friction is:
a) $E=F_n\mu \Delta x = mgcos\theta \mu \frac{25}{sin\theta}=\frac{25mg\mu}{tan\theta}$
Thus, we find:
$\frac{1}{2}mv^2 = -\frac{25mg\mu}{tan\theta}+mgh$
$v= \sqrt{2(-\frac{25g\mu}{tan\theta}+gh)}$
$v=20.1\ m/s$
b) We use a similar process:
$E=F_n\mu \Delta x = mgcos\theta \mu \frac{38}{sin\theta}=\frac{38mg\mu}{tan\theta}$
Thus, we find:
$\frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 = -\frac{38mg\mu}{tan\theta}+mgh$
$v= \sqrt{2(-\frac{38g\mu}{tan\theta}+gh)+v_0^2}$
$v = 30.4 \ m/s$
