Question

The man who with 55 kg on a skateboard. He had an initial speed of 2.77 m/s when descending the stairs at height of 2m . Calculate the man's potential energy when his speed is 6 m/s.

Answer 1

$mv_0^2/2+mgh_0=mv^2/2+mgh\to$

$mgh=PE=mv_0^2/2+mgh_0-mv^2/2=$

$=m(v_0^2/2+gh_0-v^2/2)=55\cdot(2.77^2/2+9.8\cdot 2-6^2/2)\approx300(J)$ . Answer