What happens to the kinetic energy when the velocity of the body is doubled at constant mass?

As known the expression, $K=\frac{1}{2}mv^2$

Where $m\rightarrow$ mass of the object

$v\rightarrow$ velocity of the object.

If the velocity is doubled, it becomes $2v$

The kinetic energy $K'=\frac{1}{2}m(2v)^2$

$K'=\frac{1}{2}m\times4v^2$

Or $K'=4\times\frac{1}{2}mv^2$

Or $K'=4K$

Therefore, when the velocity of the object is doubled, the kinetic energy of the object becomes four times.

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Updated on: 10-Oct-2022

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