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Question

A 1000 kilogram roller coaster goes down a hill that is 90 meters tall. Its speed at the bottom is 40 m/sec.

a. What is the efficiency of the roller coaster? Assume it starts from rest at the top of the hill.
b. What do you think happens to the “lost” energy?
c. Use the concepts of energy and efficiency to explain why the first hill on a roller coaster is the tallest.

Answer 1

Potential energy at top of hill = m*g*h = 1000*9.81*90 = 882,900 Joules = 883kJ (corrected to 3 significant figures)

 

Kinetic energy at foot of hill = 1/2 mv^2 = 1/2*1000 (40)^2 = 800,000 J

 

efficiency of energy transfer = output/input*100% = 800,000/883,000 *100% = 90.6% (Answer)

 

Note that the "lost" energy during the journey is largely due to friction along the road and air resistance.