JEE Main202523 Jan 2025Morning ShiftPhysicsWork, Power and EnergyActual
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 Joule (round off to the nearest integer)
Correct answer
0
Step-by-step solution
aligned & y=10-x & w= ₀^4 x^2(10-x) d x+ ₀^2 y^2 d y & = 10 x^3 3 - . x^4 4 |₀ ^4+ . y^3 3 |₀ ^2 & = 640 3 - 256 4 + 8 3 & =216 64 & =152 ~J aligned