## Essential University Physics: Volume 1 (3rd Edition)

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: $\frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 = -\frac{38mg\mu}{tan\theta}+mgh$ $v= \sqrt{2(-\frac{38g\mu}{tan\theta}+gh)+v_0^2}$ $v = 30.4 \ m/s$