If the slide is frictionless, what should the person's velocity be at the bottom of the slide in meters/second?

Question

The "Niagara Falls" water slide to waterpark and has a height of 76.0 feet (23.2 meters). Because the slide is not completely vertical the total length of the slide is 90.0 feet (27.4 meters). A 70.0 kg person slides down this slide starting from rest.

a) If the slide is frictionless, what should the person's velocity be at the bottom of the slide in meters/second?

b) What is the velocity in miles per hour?

c)What if there was friction between the slide and the person. If the velocity of the person at the bottom of the slide was really 38.2 mi/hour how much energy was converted to Thermal Energy in this problem?

d) From what you founded above, What must the magnitude of the friction force have been? (friction force remained the same and constant)

Please do the calculation for me I I got stuck in this problem. Thank you so much.

Answer 1

a) Velocity = sqrt(2gh) = 21.3 m/s

b) 21.3 m/s * 3.28 f/m*3600 sec/hr/*1 mile/5280 feet = 47.7 mph

c) Etop = mgh = 70*9.81*23.2 = 15931

Ebot = 1/2mv^2 = 1/2*70*21.3^2 = 10212

Eloss = Etop - Ebot = 5719 Joules

d) friction force * distance = Eloss. Thus f= 5719/90=63 Newtons