Question

A roller coaster begins at rest at the top of a 38 m hill. Find the velocity at a point 8 m above the bottom of the hill.

Hint: PE at point 1 = (KE+PE) at point 2

Answer 1

Using Energy conservation:

KE1 + PE1 = KE2 + PE2

KE1 = initial kinetic energy = 0, Since roller coaster begins at rest

KE2 = (1/2)*m*V2^2

V2 = Speed of roller coaster at point 2 = ?

PE1 = initial potential energy = m*g*h1

PE2 = final potential energy = m*g*h2

So,

0 + m*g*h1 = (1/2)*m*V2^2 + m*g*h2

m will cancel out on both sides, So

V2 = sqrt (2*g*(h1 - h2))

Using given values:

V2 = sqrt (2*9.81*(38 - 8))

V2 = 24.26 m/sec

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