JEE Main Physics question
The potential energy of a particle of mass 5 kg moving in the x–y plane is given by U = (−7x+24y) J, x and y being in metre. Initially at t=0, the particle is at the origin (0,0) moving with a velocity of (2.4î+0.7ĵ) ms⁻¹. The magnitude of force on the particle is
Answer: (A)
Solution
= −∂U/∂x = 7, = −∂U/∂y = −24. |F| = √(7²+24²) = √(49+576) = √625 = 25.