Question

A roller coaster reaches the top of the steepest hill with a speed of 6.80 km/h. It then descends the hill, which is at an average angle of 35° and is 56.0 m long. What will its speed be when it reaches the bottom? Assume µk = 0.16.

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Mass of the roller coaster = M

Speed of the roller coaster at the top of the hill = V₁ = 6.8 km/hr = 6.8 x (1000/3600) m/s = 1.889 m/s

Average angle at which the hill descends = = 35^o

Length of the hill descend = L = 56 m

Height of the hill = H

H = LSin

Normal force on the roller coaster from the hill on the incline = N

N = MgCos

Coefficient of kinetic friction between the incline and the roller coaster = _k = 0.16

Friction force on the roller coaster = f

f = _kN

f = _kMgCos

Speed of the roller coaster when it reaches the bottom = V₂

By conservation of energy the potential and kinetic energy of the roller coaster at the top of the hill is equal to the kinetic energy of the roller coaster at the bottom of the hill plus the work done against friction.

MgH + MV₁²/2 = MV₂²/2 + fL

MgLSin + MV₁²/2 = MV₂²/2 + _kMgCosL

gLSin + V₁²/2 = V₂²/2 + _kgLCos

(9.81)(56)Sin(35) + (1.889)²/2 = V₂²/2 + (0.16)(9.81)(56)Cos(35)

V₂ = 22.13 m/s

Speed of the roller coaster when it reaches the bottom = 22.13 m/s