y=10−xw=∫04x2(10−x)dx+∫02y2dy=310x3−4x404+3y302=3640−4256+38=216×64=152 J
A force f= x^2 y i+ y^2 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…
JEE Main 2025 — Physics Mechanics
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)
Held on 23 Jan 2025 · Verified 6 Jul 2026.
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