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 61.0 kg, and the height of the water slide is 12.3 m. If the kinetic frictional force does -5.80 × 103 J of work, how fast is the student going at the bottom of the slide?

Answer 1

Answer:

Velocity will be equal to 7.31 m/sec

Explanation:

We have given mass of the student m = 61 kg

Height of the water slide h = 12.3 m

Acceleration due to gravity $g=9.8m/sec^2$

Potential energy is equal to $U=mgh=61\times 9.8\times 12.3=7352.94J$

Work done due to friction = -5800 J

So energy remained = 7352.94-5800 = 1552.94 J

This energy will be equal to kinetic energy

So $\frac{1}{2}mv^2=1552.94$

$\frac{1}{2}\times 61\times v^2=1552.94$

$v^2=50.91$

v = 7.13 m/sec