JEE MAIN - Physics (2023 - 1st February Evening Shift - No. 25)
A force $$\mathrm{F}=\left(5+3 y^{2}\right)$$ acts on a particle in the $$y$$-direction, where $$\mathrm{F}$$ is in newton and $$y$$ is in meter. The work done by the force during a displacement from $$y=2 \mathrm{~m}$$ to $$y=5 \mathrm{~m}$$ is ___________ J.
Answer
132
Explanation
$\begin{aligned} & W=\int F d y=\int_2^5\left(5+3 y^2\right) d y \\\\ & =\left.\left(5 y+y^3\right)\right|_2 ^5 \\\\ & =(15+125-8) \mathrm{J} \\\\ & =132 \mathrm{~J}\end{aligned}$
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