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

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Work, Power & Energy

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)

(4 m, 2 m)

____

Joule (round off to the nearest integer)

Answer: 152

The work done is

W = \int F \cdot d r

with

F = x^{2} y \hat{i} + y^{2} \hat{j}

, integrating the

x

-component along

x

from 0 to 4 (using

y = 10 - x

on the given plane) and the

y

-component along

y

from 0 to 2:

W = \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{y ^{3}}{3}]_{0}^{2} = \frac{8}{3}

Total work:

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

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Integer answer