Question

A 77.0 kg woman slides down a 42.6 m long waterslide inclined at 42.3. at the bottom, she is moving 20.3 m/s.how much work did friction do on the woman?(unit=J)PLEASE HELP

Answer 1

Answer:

5.791244495 KNm

Explanation:

The height h is given by, $h=42.6sin42.3^{o}$

Potential energy, PE is given by

PE=mgh where m is mass of the woman, g is acceleration due to gravity whose value is taken as $9.81 m/s^{2}$ and h is already given hence substituting 77 Kg for m we obtain

$PE=77*9.81*42.6sin42.3^{o}=21656.7095 Nm$

PE=21.6567095 KNm

We also know that Kinetic energy is given by$0.5mv^{2}$ where v is the velocity and substituting v for 20.3 we obtain

$KE=0.5*77*20.3^{2}=15865.465 Nm$

KE=15.865465 KNm

Friction work is the difference between PE and KE hence

Friction work=21.6567095 KNm-15865.465 Nm=5.791244495 KNm