Question

A roller coaster reaches the top of the steepest hill with a speed of 6.8 km/h . It then descends the hill, which is at an average angle of 35 ∘ and is 45.0 m long. Q: What will its speed be when it reaches the bottom? Assume μk = 0.18.

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Mass of the roller coaster = m

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

Angle of incline the roller coaster descends at = = 35^o

Length of the descend = L = 45 m

Height of the descend = H

H = LSin

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

Coefficient of kinetic friction = _k = 0.18

Friction force on the roller coaster = f

f = _kN

f = _kmgCos

Speed of the roller coaster at the bottom = V₂

The initial potential and kinetic energy of the roller coaster is equal to the kinetic energy of the roller coaster at the bottom 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)(45)Sin(35) + (1.889)²/2 = V₂²/2 + (0.18)(9.81)(45)Cos(35)

V₂ = 19.5 m/s

Speed of the roller coaster at the bottom of the hill = 19.5 m/s