Work Energy Power

JEE Main level, numerical answer, medium

7 of 183

Question 7 of 183

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)

Show answer and solution

Answer: 152

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}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Collisions, centre of mass and rockets simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Work Energy Power (Work done), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.