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Q. A force $F =\left(5+3 y^2\right)$ acts on a particle in the $y$-direction, where $F$ is in newton and $y$ is in meter. The work done by the force during a displacement from $y=2 m$ to $y=5 m$ is ___-$J$.

JEE MainJEE Main 2023Work, Energy and Power

Solution:

$ F =5+3 y ^2 $
$ W=\int\limits_2^5\left(5+3 y^2\right) d y $
$=\left[5 y+\frac{3 y^3}{3}\right]_2^5 $
$ =132\, J$