Work Energy Power

JEE Main level, single correct, medium

62 of 183

Question 62 of 183

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:

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Answer: (C) 44

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{matrix}{F}_{x} & = & -\dfrac{\partial U}{\partial x} \\ & = & -4x\end{matrix}
Hence, at the point (1,2,3)m\left(1,2,3\right)m, the value of the force is given by
∣Fx∣=∣−4×1∣N=4N\begin{matrix}\left|{F}_{x}\right| & = & \left|-4\times 1\right|N \\ & = & 4N\end{matrix}

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