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JEE Main 2021 (25 Jul, Shift 2) · Work by a Variable Force

Work by a Variable Force question from JEE Main 2021 (25 Jul, Shift 2)

Q13JEE Main 2021 · 25 JUL · SHIFT 2Easy

Work by a Variable Force

A force of F = (5y + 20) $\hat{j}$ N acts on a particle. The work done by this force when the particle is moved from y = 0 m to y = 10 m is ___________ J.

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

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Answer: 450

$W = \int_0^{10} (5y + 20)\,dy = \left[\frac{5y^2}{2} + 20y\right]_0^{10}$
$W = 250 + 200$
Work done $= 450$ J.

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