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 74.0 kg, and the height of the water slide is 11.3 m. If the kinetic frictional force does -5.42 × 103 J of work, how fast is the student going at the bottom of the slide?

Answer 1

Answer:

The student is going at the bottom of the slide with a velocity of 8.66 m/s

Explanation:

Given;

mass of the student, m = 74 kg

height of the water slide, h = 11.3 m

work done, W = -5.42 × 10³ J

Apply work energy theorem;

$W = \frac{1}{2}mv_f^2+ mgh_f-(\frac{1}{2}mv_o^2 + mgh_o)\\\\ W + \frac{1}{2}mv_o^2 + mgh_o -mgh_f= \frac{1}{2}mv_f^2\\\\ W + \frac{1}{2}mv_o^2 +mg(h_o-h_f) = \frac{1}{2}mv_f^2\\\\\frac{W}{m} + \frac{1}{2} v_o^2 + g(h_o-h_f) = \frac{1}{2}v_f^2\\\\\frac{2W}{m}+ v_o^2 + 2g(h_o-h_f) = v_f^2\\\\v_f = \sqrt{\frac{2W}{m}+ v_o^2 - 2g(h_f-h_o)} \\\\v_f = \sqrt{\frac{2(-5.42*10^3)}{74}+ (0)^2 - 2*9.8(-11.3)}\\\\v_f=\sqrt{-146.4865+221.48} \\\\v_f = \sqrt{74.9935} \\\\v_f = 8.66 \ m/s$

Therefore, the student is going at the bottom of the slide with a velocity of 8.66 m/s