A body of $2\ kg$ falls from rest. What will be its kinetic energy during the fall at the end of $2\ s$? $(Assume\ g=10\ m/s^2)$

Here given, mass of the body $m=2\ kg$

Initial velocity $u=0$

Gravitational acceleration $g=10\ m/s^2$ 

Time $t=2\ s$

On using the first equation of the motion $v=u+gt$

$v=0+10\times2$

Or $v=20\ m/s$

Therefore, the kinetic energy of the body $K=\frac{1}{2}mv^2$

$=\frac{1}{2}\times2\times20^2$

$=400\ J$

Therefore, the kinetic energy of the body at the end of $2\ s$ will be $400\ Joule$.

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Updated on: 10-Oct-2022

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