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

<p>The force acting on a particle can be determined from the potential energy function by using the negative gradient. In three-dimensional space, the force vector <strong>F</strong> is related to the potential energy function $ U $ by:</p> <p>$\mathbf{F} = -\nabla U$</p> <p>where $ \nabla U $ (the gradient of $ U $) is a vector with components given by the partial derivatives of $ U $ with respect to $ x $, $ y $, and $ z $:</p> <p>$$ \nabla U = \left( \frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}, \frac{\partial U}{\partial z} \right) $$</p> <p>Given the potential energy function:</p> <p>$U = 2x^2 + 3y^3 + 2z$</p> <p>Calculate the partial derivative of $ U $ with respect to $ x $ to find the $ x $-component of the force:</p> <p><strong>Partial Derivative with respect to $ x $:</strong></p> <p>$$ \frac{\partial U}{\partial x} = \frac{\partial}{\partial x}(2x^2 + 3y^3 + 2z) = 4x $$</p> <ol start="2"> <p><strong>Calculate the $ x $-component of the force:</strong></p> <p>Since $ F_x = -\frac{\partial U}{\partial x} $, the $ x $-component of the force is:</p> <p>$F_x = -4x$</p> <ol start="3"> <p><strong>Evaluate at point $ P(1,2,3) $:</strong></p> <p>Substitute $ x = 1 $ into the expression for $ F_x $:</p> <p>$F_x = -4(1) = -4$</p> <p>The magnitude of the $ x $-component of the force is:</p> <p>$|F_x| = 4 \, \text{N}$</p> <p>Thus, the correct answer is <strong>Option A: 4</strong>.</p>