Question

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

• A. 2
• B. 6
• C. 4
• D. 8

Answer 1

Answer: (C)

4

Answer 2

Step 1: Given Potential Energy Function

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

Step 2: Calculate the $x$-Component of Force

The force in the $x$-direction is given by $F_x = -\frac{\partial U}{\partial x}$. Differentiating $U$ with respect to $x$:

$F_x = -\frac{\partial}{\partial x}(2x^2) = -4x$

Step 3: Evaluate $F_x$ at $x = 1$

Substitute $x = 1$:

$F_x = -4 \times 1 = -4$

The magnitude of $F_x$ is 4 N.

So, the correct answer is: 4 N