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 (integer)
132
Solution
$\begin{aligned} & W=\int F d y=\int_2^5\left(5+3 y^2\right) d y \\\\ & =\left.\left(5 y+y^3\right)\right|_2 ^5 \\\\ & =(15+125-8) \mathrm{J} \\\\ & =132 \mathrm{~J}\end{aligned}$
About this question
Subject: Physics · Chapter: Work, Energy and Power · Topic: Work Done by a Force
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