Conservation of Energy - MCAT Chemical and Physical Foundations of Biological Systems

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Question

A ball with mass of 2kg is dropped from the top of a building this is 30m high. What is the approximate velocity of the ball when it is 10m above the ground?

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Answer

Use conservation of energy. The gravitational potential energy lost as the ball drops from 30m to 10m equals the kinetic energy gained.

Change in gravitational potential energy can be found using the difference in mgh.

So 400 Joules are converted from gravitational potential to kinetic energy, allowing us to solve for the velocity, v.

400 Joules = \frac{1}{2}mv^{2}

400 = v^{2}

v = 20 m/s

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