Question

In: Physics

A roller coaster reaches the top of the steepest hill with a speed of 6.8 km/h...

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.

Solutions

Expert Solution

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 = = 35o

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


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