A block of mass $5\ kg$ is lying on a frictionless table. A force of $20\ N$ is applied on it for $10$ seconds. Calculate its kinetic energy.

Here given, mass of the block $m=5\ kg.$, Applied force $F=20\ N$, Time $t=10\ sec.$, Initial velocity $u=0$

Let $a$ be the acceleration of the block.

Therefore, $a=\frac{F}{m}$

Or $a=\frac{20\ N}{5\ kg}$

Or $a=4\ ms^{-2}$

On using the equation of motion $v=u+at$

$v=0+4\times10$

Or $v=40\ ms^{-1}$

Therefore, Kinetic energy $K=\frac{1}{2}mv^2$

$=\frac{1}{2}\times5\times(40)^2$

$=4000\ Joule$

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Updated on: 10-Oct-2022

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