Question

A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force) acts on her. The student has a mass of 80.0 kg, and the height of the water slide is 12.5 m. If the kinetic frictional force does -5.10 103 J of work, how fast is the student going at the bottom of the slide

Answer 1

v= 11.06 m/s

Step-by-step explanation:

Given that

Mass of the student ,m = 80 kg

Height ,h= 12.5 m

Force done by friction ,W'=-5.1 x 10³ J

Lets take the sped of the student become v m/s at the bottom.

From work power energy

Work done by all forces = Change in the kinetic energy

Work done by Gravity + Work done by friction = ΔKE

Initially the speed of the student is zero

`m g h + W' = (1)/(2)mv^2`

m v² =2 m g h + 2 W'

80 v² =2 x 80 x 10 x 12.5 - 2 x 5.1 x 10³

80 v² =9800

v= 11.06 m/s