QJEEM24Ph8ac9: JEE Main 2024 Physics - Class 11 JEEM24Ph8ac9 Solution: The potential energy function (in ) of a particle in a region…

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 F is related to the potential energy function $U$ by:

$\mathbf{F} = -\nabla U$

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 $z$ :

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

Given the potential energy function:

$U = 2x^2 + 3y^3 + 2z$

Calculate the partial derivative of $U$ with respect to $x$ to find the $x$ -component of the force:

Partial Derivative with respect to $x$ :

$\frac{\partial U}{\partial x} = \frac{\partial}{\partial x}(2x^2 + 3y^3 + 2z) = 4x$

Calculate the $x$ -component of the force:

Since $F_x = -\frac{\partial U}{\partial x}$ , the $x$ -component of the force is:

$F_x = -4x$

Evaluate at point $P(1,2,3)$ :

Substitute $x = 1$ into the expression for $F_x$ :

$F_x = -4(1) = -4$

The magnitude of the $x$ -component of the force is:

$|F_x| = 4 \, \text{N}$

Thus, the correct answer is Option A: 4.