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

-5770

Step-by-step explanation:

put it in acellus and it was right!

Answer 2

5.791244495 KNm

Step-by-step 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