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

The Correct Option is C

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