JEE Main 2025 Jan 23 Shift 1, Physics Q24: Work Energy Theorem And Its Applications

JEE Main2025Jan 23, Shift 1Physics

A force $f = x^{2} y \hat{i} + y^{2} \hat{j}$ acts on a particle in a plane $x + y = 10$ . The work done by this force during a displacement from $(0, 0)$ to $(4 m, 2 m)$ is _____ Joule (round off to the nearest integer)

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Solution

Since the particle moves in the plane $x + y = 10$ , we have $y = 10 - x$ along the $\hat{i}$ -component of $f$ . The work integral splits:

$W = \int F_{x} d x + \int F_{y} d y = \int_{0}^{4} x^{2} (10 - x) d x + \int_{0}^{2} y^{2} d y$ .

First integral: $\int_{0}^{4} (10 x^{2} - x^{3}) d x = [\frac{10 x ^{3}}{3} - \frac{x ^{4}}{4}]_{0}^{4} = \frac{640}{3} - 64 = \frac{448}{3}$ .

Second integral: $\int_{0}^{2} y^{2} d y = \frac{8}{3}$ .

$W = \frac{448 + 8}{3} = \frac{456}{3} = 152$ J.

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JEE Main 2025 Jan 23 Shift 1, Physics Q24: Work Energy Theorem And Its Applications

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