A force (3 x 2 +2 x-5 ) N displaces a body from x =2 m to x =4 m . Work done by this force is ............J.

Physics · Work, Power & Energy

JEE Main 2024 — 9 April, Shift 2 — Question 54

A force $\left(3 x^{2}+2 x-5\right) N$ displaces a body from $\mathrm{x}=2 \mathrm{~m}$ to $\mathrm{x}=4 \mathrm{~m}$ . Work done by this force is ............J.

Answer: 58

Numerical answer — enter this value.

Step-by-step solution

$\quad \mathrm{W}=\int^{\mathrm{x}_{2}} \mathrm{Fdx}$

$W=\int_{2}^{4}\left(3 x^{2}+2 x-5\right) d x$

$\mathrm{W}=\left[\mathrm{x}^{3}+\mathrm{x}^{2}-5 \mathrm{x}\right]_{2}^{4}$

$\mathrm{W}=[60-2] \mathrm{J}=58 \mathrm{~J}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 2
Subject: Physics
Chapter: Work, Power & Energy
Topic: Work Done by a Constant and Variable Force