Question #123214

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)

Expert's answer

**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)}."

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