JEE Main202125 Jul 2021Evening ShiftPhysicsWork, Power and EnergyActual
A force of F = ( 5 y + 20 ) j ^ N acts on a particle. The work done by this force when the particle is moved from y = 0 m to y = 10 m is ________ J .
Correct answer
0
Step-by-step solution
F = ( 5 y + 20 ) j ^ ω = ∫ F d y = ∫ 0 10 ( 5 y + 20 ) dy = 5 y 2 2 + 20 y 0 10 = 5 2 × 100 + 20 × 10 = 250 + 200 = 450   J