Question

The Outlaw Run roller coaster in Branson, Missouri, features a track that is inclined at 84 ∘ below the horizontal and that spans a 49.4-m (162 ft) change in height.
a)If the coefficient of friction for the roller coaster is μk = 0.28, what is the magnitude of the friction force on a 1600-kg coaster train?
b)How much work is done by friction during the time the coaster train travels along the incline?
c)What is the speed of the coaster train at the bottom of the incline, assuming it started from rest at the top?

Answer 1

Answer:

Explanation:

a)

Ff = μmgcosθ

Ff = 0.28(1600)(9.8)cos(-84)

Ff = 458.9217...

Ff = 460 N

b)  ignoring the curves required at top and bottom which change the friction force significantly, especially at the bottom where centripetal acceleration will greatly increase normal forces and thus friction force.

W = Ffd

W = 458.9217(-49.4/sin(-84)

W = 22,795.6119...

W = 23 kJ

c) same assumptions as part b

The change in potential energy minus the work of friction will be kinetic energy.

KE = PE - W

½mv² = mgh - (μmgcosθ)d

v² = 2(gh - (μgcosθ)(h/sinθ))

v = √(2gh(1 - μcotθ))

v = √(2(9.8)(49.4)(1 - 0.28cot84))

v = 30.6552...

v = 31 m/s