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 creat am LOL to help solve the problem)

Answer 1

Solution.

Initially, the roller car had potential energy, which turned into kinetic and work against friction forces.

$E_p=W_f+E_k;$

$mgh=\mu mgx+\dfrac{mv^2}{2},$ x is the distance at which the friction force acted.

$\dfrac{v^2}{2}=gh-\mu gx;$

$v^2=2(gh-\mu gx);$

$v=\sqrt{2(gh-\mu gx)}=\sqrt{2g(h-\mu x)};$

Answer: $v=\sqrt{2g(h-\mu x)}.$