CLASSICAL MECHANICS & THERMODYNAMICS - Chapter 3, Exercise 2 Solution
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Speed from Conservation of Energy on a Ramp

PROBLEM
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A 0.5 kg skateboard-and-rider system (highly simplified as a point
mass) starts at rest at the top of a 3 m high, frictionless ramp.
Using conservation of energy, find the speed at the bottom of the
ramp.

SOLUTION
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At the top of the ramp, the system is at rest, so it has zero kinetic
energy and maximum potential energy (relative to the bottom of the
ramp). At the bottom, all of that potential energy has converted into
kinetic energy (the ramp is frictionless, so no energy is lost to
heat).

    PE (top) = KE (bottom)
    mgh = 1/2 m v^2

Mass appears on both sides and cancels:

    gh = 1/2 v^2
    v^2 = 2gh
    v^2 = 2 x 9.81 x 3
    v^2 = 58.86
    v = sqrt(58.86)
    v = 7.67 m/s (approximately)

ANSWER: The system reaches a speed of approximately 7.67 m/s at the
bottom of the ramp.

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WHY THIS WORKS AS AN ANSWER

Because the ramp is stated to be frictionless, no mechanical energy
is lost to heat along the way, so the total mechanical energy (PE +
KE) is conserved from top to bottom - exactly the same setup as the
chapter's own worked example of a ball dropped from a height. Mass
cancels out of the equation entirely, which is why the given 0.5 kg
value never actually needed to be substituted in - it does not affect
the final speed at all, only the total amount of energy involved. This
is the same real physical reason a heavier and a lighter object reach
the same speed falling (or rolling down a frictionless slope) from
the same height.