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MCQEasyJEE 2024Potential Energy & Conservative Forces

JEE Physics 2024 Question with Solution

The potential energy function (in J) of a particle in a region of space is given as U=(2x2+3y3+2z)U = (2x^2 + 3y^3 + 2z). Here xx, yy, and zz are in meters. The magnitude of the xx-component of force (in N) acting on the particle at point P(1,2,3)P(1, 2, 3) m is:

  • A

    22

  • B

    66

  • C

    44

  • D

    88

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The potential energy function is U=2x2+3y3+2zU = 2x^2 + 3y^3 + 2z and the point is P(1,2,3)P(1,2,3).

Find: The magnitude of the xx-component of force.

The force is related to potential energy by

F⃗=−∇U\vec{F} = -\nabla U

So, the xx-component is

Fx=−∂U∂xF_x = -\frac{\partial U}{\partial x}

Differentiate UU with respect to xx:

∂U∂x=∂∂x(2x2)=4x\frac{\partial U}{\partial x} = \frac{\partial}{\partial x}(2x^2) = 4x

Hence,

Fx=−4xF_x = -4x

At x=1x = 1,

Fx=−4(1)=−4 NF_x = -4(1) = -4 \, \text{N}

Therefore,

∣Fx∣=4 N|F_x| = 4 \, \text{N}

Therefore, the magnitude of the xx-component of force is 4 N4 \, \text{N}. The correct option is C.

Gradient-Based Expansion

Given: U=2x2+3y3+2zU = 2x^2 + 3y^3 + 2z.

Find: Magnitude of the xx-component of force at P(1,2,3)P(1,2,3).

In three dimensions,

∇U=(∂U∂x,∂U∂y,∂U∂z)\nabla U = \left( \frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}, \frac{\partial U}{\partial z} \right)

Now calculate the partial derivatives:

∂U∂x=4x\frac{\partial U}{\partial x} = 4x ∂U∂y=9y2\frac{\partial U}{\partial y} = 9y^2 ∂U∂z=2\frac{\partial U}{\partial z} = 2

So the force vector is

F⃗=−(4x,9y2,2)\vec{F} = -(4x, 9y^2, 2)

Hence the xx-component is

Fx=−4xF_x = -4x

Substituting x=1x = 1,

Fx=−4F_x = -4

Its magnitude is 44. Therefore, the answer is 4 N4 \, \text{N}.

Common mistakes

  • Using Fx=∂U∂xF_x = \frac{\partial U}{\partial x} instead of Fx=−∂U∂xF_x = -\frac{\partial U}{\partial x}. Force is the negative gradient of potential energy, so the sign must be negative before taking magnitude.

  • Substituting all coordinates into the full force vector even though only the xx-component is asked. Here only differentiation with respect to xx is needed to find FxF_x.

  • Reporting −4 N-4 \, \text{N} as the final answer. The question asks for the magnitude of the xx-component, so the correct final value is 4 N4 \, \text{N}.

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