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Work Power and Energy question

2025 · 23 Jan · Shift 1 · Q72
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Work Power and Energy question

2025 · 23 Jan · Shift 1 · Q72

JEE MainPhysicsWork Power and EnergyNumerical+4 / −1
A force f=x2yi^+y2j^\mathrm{f}=\mathrm{x}^2 \mathrm{y} \hat{\mathrm{i}}+\mathrm{y}^2 \hat{\mathrm{j}}f=x2yi^+y2j^​ acts on a particle in a plane x+y=10\mathrm{x}+\mathrm{y}=10x+y=10. The work done by this force during a displacement from (0,0)(0,0)(0,0) to (4 m,2 m)(4 \mathrm{~m}, 2 \mathrm{~m})(4 m,2 m) is ‾\underline{\hspace{2cm}}​ Joule (round off to the nearest integer)
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