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JEE Main 202523 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Morning Shift
Subject
Physics
Chapter
Work, Power and Energy

JEE Main 2025 · 23 January · Morning Shift · Physics · Work, Power and Energy

A force f=x^2 y i +y^2 j acts on a particle in a plane x+y=10 . The...

A force f=x2yi^+y2j^\mathrm{f}=\mathrm{x}^2 \mathrm{y} \hat{\mathrm{i}}+\mathrm{y}^2 \hat{\mathrm{j}} acts on a particle in a plane x+y=10\mathrm{x}+\mathrm{y}=10. The work done by this force during a displacement from (0,0)(0,0) to (4 m,2 m)(4 \mathrm{~m}, 2 \mathrm{~m}) is \qquad Joule (round off to the nearest integer)

Correct answer

0

Step-by-step solution

y=10−xw=∫04x2(10−x)dx+∫02y2dy=10x33−x44∣04+y33∣02=6403−2564+83=216×64=152 J\begin{aligned} & y=10-x \\ & w=\int_0^4 x^2(10-x) d x+\int_0^2 y^2 d y \\ & =\frac{10 x^3}{3}-\left.\frac{x^4}{4}\right|_0 ^4+\left.\frac{y^3}{3}\right|_0 ^2 \\ & =\frac{640}{3}-\frac{256}{4}+\frac{8}{3} \\ & =216 \times 64 \\ & =152 \mathrm{~J}\end{aligned}
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