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JEE Main 2023 (1 Feb, Shift 2) · Work by a Variable Force

Work by a Variable Force question from JEE Main 2023 (1 Feb, Shift 2)

Q6JEE Main 2023 · 1 FEB · SHIFT 2Easy

Work by a Variable Force

A force $F = \left(5 + 3y^2\right)$ 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 $y = 2\text{ m}$ to $y = 5\text{ m}$ is ___________ J.

Numerical answer type. Work it out, then open the answer.

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Answer: 132

$W = \int_2^5 (5 + 3y^2)\,dy = \left[5y + y^3\right]_2^5$
$W = 5(3) + (125 - 8) = 15 + 117$
Work done $= 132$ J.

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