Verified Solution • 2025–26 Session moderate

QJEEM23Ph05b8: JEE Main 2023 Physics - Class 11 JEEM23Ph05b8 Solution: A force acts on a particle in the -direction, where is in…

A force F=(5+3y2)\mathrm{F}=\left(5+3 y^{2}\right) acts on a particle in the yy -direction, where F\mathrm{F} is in newton and yy is in meter. The work done by the force during a displacement from y=2 my=2 \mathrm{~m} to y=5 my=5 \mathrm{~m} is ___________ J.

Step-by-Step Verified Solution

amp;W=∫Fdy=∫25(5+3y2)dyamp;=(5y+y3)∣25amp;=(15+125−8)Jamp;=132 J\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}

Related Learning Resources & Exam Preparation