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 J

Step-by-step explanation:

Potential energy = kinetic energy + work

PE = KE + W

mgh = ½ mv² + W

(77.0 kg) (9.8 m/s²) (42.6 m sin 42.3°) = ½ (77.0 kg) (20.3 m/s)² + W

W = 5770 J