A cart starts at rest and moves freely down a ramp without friction or air resistance and descends 8 meters vertically, what is its speed at the bottom?

We apply energy conservation. At the start the cart has only gravitational potential energy given by mgh where m is its mass, g is the gravitational field, h = 8m is the height. At the end the cart has only kinetic energy mv^2/2 where v is its speed. By conservation of energy mgh = mv^2/2, so v^2=2gh= 29.88 m^2 s^-2=156.8 m^2 s^-2 so taking the square root, v = 12.5 m s^-1.

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Answered by Joshua T. Physics tutor

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