The potential energy function (in J ) of a particle in a region of space is given as U=(2 x²+3 y³+2 z). 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 :

Solution

Physics sir · 3 steps

Force is minus the slope of U, so differentiate with respect to x and then put in x = 1.

1. Force from U. Each force component is the negative partial derivative of U along that axis. For the x-component, treat y and z as constants.

   Fₓ = -(∂ U)/(∂ x) = -∂/(∂ x)(2x² + 3y³ + 2z) = -4x

2. Put in the point. At P(1, 2, 3) only x = 1 matters, so Fₓ = -4 × 1 = -4 N. The minus sign means the force points along -x.
3. So. The magnitude of the x-component of force is 4 N, option (a).

The idea to keep. To find a force component from U, differentiate U only along that axis, with the other coordinates held fixed. Substitute the point only after you have differentiated.

Answer from the NTA final answer key. Solution by PrepWiser.