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 velocity (2.4î+0.7ĵ) ms⁻¹. The magnitude of the force on the particle is
Answer: (A)
Solution
(Same as Q217.) **F** = 7î − 24ĵ; |**F**| = √(7²+24²) = 25 units.