Answer to Question #116306 in Mechanics | Relativity for Damian

Question

When the roller derby car reaches the bottom of the hill. How fast is it traveling (point C)? (using g = 10m/s/s and create an LOL to help solve the problem)

Answer 1

From the LOL diagram above, we see that the potential energy turned into work done against friction and kinetic energy of derby car motion:

$$PE=FR+KE,\\ mgh=\mu mgx+\frac{mv^2}{2},\\ \space\\ v=\sqrt{2g(h-\mu x)}.$$

Just substitute the initial height $h$ , coefficient of friction $\mu$, and the distance $x$ where the friction was involved.