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(D) The kinetic energy $(K.E.)$ of an object of mass $m$ moving with velocity $v$ is given by the formula: $K.E. = \frac{1}{2}mv^2$.If the speed is do

Vedclass · Accepted Answer

kinetic energy increases $4$ times

Vedclass · Answer

(D) The kinetic energy $(K.E.)$ of an object of mass $m$ moving with velocity $v$ is given by the formula: $K.E. = \frac{1}{2}mv^2$.If the speed is doubled,the new velocity becomes $v' = 2v$.The new kinetic energy $(K.E.')$ will be: $K.E.' = \frac{1}{2}m(2v)^2 = \frac{1}{2}m(4v^2) = 4 	imes (\frac{1}{2}mv^2)$.Therefore,$K.E.' = 4 	imes K.E.$This shows that the kinetic energy increases $4$ times when the speed is doubled.

When the speed of a moving object is doubled,its:

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