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.
"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
Comments
Leave a comment