The Outlaw Run roller coaster in Branson, Missouri, features a track that is inc

Question

The Outlaw Run roller coaster in Branson, Missouri, features a track that is inclined at 77 below the horizontal and that spans a 49.4-m(162 ft) change in height.

If the coefficient of friction for the roller coaster is k=0.31,what is the magnitude of the friction force on a 1600-kg coaster train?

How much work is done by friction during the time the coaster train travels along the incline?

What is the speed of the coaster train at the bottom of the incline, assuming it started from rest at the top?

Explanation / Answer

theta = 77 deg below horizontal
dh = 49.4 m
k = 0.31
m = 1600 kg
friction force = kmgcos(theta) = 1094.5578 N
work done by friction = -f*dh/sin(77) = -55493.45 J
work done by gravity = mgdh = 775382.4 J
final KE of train = work done by friction + work done by gravity = 719888.95 J
final speed = v
0.5mv^2 = final KE
v = 29.997 m/s

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