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

Question

Please help me by solving this problem for me. Thank you so much I appreciate it a lot.

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 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?

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

Answer 1

(a) Use conservation of mechanical energy. At top of slide person has potential energy mgh.

At bottom of slide all this has changed to kinetic energy 1/2 mv²

So mgh = 1/2 mv². m = 70, g=9.81, h = 23.2 so you can find v

(b) 1 mile = 1.609 km = 1609 m, 1 hour = 3600 s. Convert 38.2 mi/hr to m/s. Using this find kinetic energy at bottom. Energy converted to thermal = potential - kinetic

(c) Force x distance = work done = energy change. So as slide is 27.4 m long, divide the thermal energy from (b) by the 27.4 to get this