A force f = x²yi + y²j 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) Numerical answer type. Work it out, then open the answer.

Answer: 152. W = ∫ fx dx + ∫ fy dy = ∫ x² y dx + ∫ y² dy Using the given constraint y = 10 - x in the first term: ∫₀⁴ x²(10 - x) dx = (10x³)/(3) - x⁴/4|₀⁴ = 640/3 - 64 ≈ 149.3 J ∫₀² y² dy = 8/3 ≈ 2.7 J W ≈ 149.3 + 2.7 = 152 J

A force f = x²yi + y²j acts on a particle in a plane x + y = 10. The… | JEE Main 2025 (23 Jan, Shift 1)

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Work by a Variable Force

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

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

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Answer: 152

$W = \int f_x\,dx + \int f_y\,dy = \int x^2 y\,dx + \int y^2\,dy$
Using the given constraint $y = 10 - x$ in the first term: $\int_0^4 x^2(10 - x)\,dx = \frac{10x^3}{3} - \frac{x^4}{4}\Big|_0^4 = \frac{640}{3} - 64 \approx 149.3\text{ J}$
$\int_0^2 y^2\,dy = \frac{8}{3} \approx 2.7\text{ J}$
$W \approx 149.3 + 2.7 = 152\text{ J}$

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