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JEE Main 202429 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
29 January
Shift
Morning Shift
Subject
Physics
Chapter
Work, Power and Energy

JEE Main 2024 · 29 January · Morning Shift · Physics · Work, Power and Energy

The potential energy function (in J ) of a particle in a region of...

The potential energy function (in JJ ) of a particle in a region of space is given as U=(2x2+3y3+2z)U = \left(2{x}^{2} + 3{y}^{3} + 2z\right). Here x,yx, y and zz are in meter. The magnitude of xx - component of force (in NN ) acting on the particle at point P(1,2,3)mP ( 1, 2, 3 ) m is:

Options

  1. A22
  2. B66
  3. C44
  4. D88

Correct answer

C. 44

Step-by-step solution

Given U=2x2+3y3+2zU = 2{x}^{2} + 3{y}^{3} + 2z

The expression for the x−x -component of the force can be calculated as follows: 

{Fx=−∂U∂x=−4x\begin{cases}{F}_{x} & = & - \dfrac{∂ U}{∂ x} \\ & = & - 4x\end{cases}

Hence, at the point \left(1, 2, 3\right)m,thevalueoftheforceisgivenby</p><p>, the value of the force is given by</p><p>\begin{cases}\left|{F}_{x}\right| & = & \left|- 4 \times 1\right|N \\ & = & 4N\end{cases}$

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