JEE Main 2024 Physics question
The potential energy function (in J) of a particle in a region of space is given as U = (2x² + 3y² + 2z). Here x, y and z are in meter. The magnitude of x component of force (in N) acting on the particle at point P(1, 2, 3)m is:
Answer: (D)
Solution
= −∂U/∂x = −4x, so at x = 1 its magnitude is 4 N.