Conservation Of Mechanical Energy

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A roller-coaster car starts from rest at the top of a smooth, frictionless hill that is 20 m high and rolls down to the bottom. What speed does the car reach at the base of the hill?

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On a frictionless track, mechanical energy is conserved, so the gravitational potential energy lost equals the kinetic energy gained: $mgh = 	frac{1}{2}mv^2$. The mass cancels, and the shape of the hill does not matter because gravity is a conservative force; only the vertical drop counts. Solving gives $v = \sqrt{2gh}$, and with $g = 10	ext{ m/s}^2$ and $h = 20	ext{ m}$ this is $v = \sqrt{2 	imes 10 	imes 20} = \sqrt{400} = 20	ext{ m/s}$. The option 10 m/s omits the factor of two inside the root. The option 400 m/s reports $v^2$ without taking the square root. The option 14 m/s corresponds to half the height, $h = 10	ext{ m}$. The cancellation of mass is significant: a heavy car and a light car released from the same height arrive at identical speeds, a direct echo of Galileo's observation that all bodies fall alike. The independence from the path's shape is exactly what lets engineers design loops and dips while still predicting the speed from the vertical drop alone. This is the NCERT principle of conservation of mechanical energy for conservative fields. As a check, the answer matches free-fall from 20 m, which it must, since the smooth track only redirects motion.

Conservation Of Mechanical Energy

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