A 1500 kg roller coaster train is at rest at the top of a 65 m drop before it begins to travel down
the hill. At the bottom of the hill how fast is the train moving?

correct  Vmax = 36 m/s

Explanation:

Given:

m = 1,500 kg  the mass of the roller coaster

H = 65 m  the height of the hilltop

Vmax = ?  the maximum speed which roller coaster will have at the bottom of the hill

We will solve this problem using energy conservation laws:

At the top of the hill, the total roller coaster energy is equal to the maximum potential energy while the kinetic energy is zero:

Etotal = Epmax  and Ek = 0 J

The formula for calculating potential energy is:

Ep = m g H   we will take  g = 10 m/s²

Epmax = m g H

At the bottom of the hill, the total roller coaster energy is equal to the maximum kinetic energy while the potential energy is zero:

Etotal = Ekmax  and Ep = 0 J

The formula for calculating kinetic energy is:

Ek = m V² / 2

Ekmax = Etotal = Epmax

Ekmax = m Vmax² / 2  = m g H

m Vmax² / 2  = m g H   when we divide both sides by mass m we get:

Vmax² / 2  = g · H  ⇒  Vmax² = 2 · g · H

Vmax² = 2 · 10 · 65 = 1300

Vmax = √1300 = 36 m/s

Vmax = 36 m/s

God is with you!!!

the answer would be c

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