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

To determine the magnitude of the \( x \)-component of the force on a particle at point \( P(1, 2, 3) \), the relationship between potential energy and force is utilized. Force components are derived from the negative gradient of the potential energy function.

The potential energy function is defined as:

\(U = (2x^2 + 3y^3 + 2z)\)

The force acting on the particle is expressed as:

\(\vec{F} = -abla U\)

In this equation, \(abla U\) denotes the gradient of the potential energy function \(U\). The three-dimensional gradient is calculated as:

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

The partial derivatives are computed as follows:

Partial derivative with respect to \(x\):

\(\frac{\partial U}{\partial x} = \frac{\partial}{\partial x}(2x^2) = 4x\)

Partial derivative with respect to \(y\):

\(\frac{\partial U}{\partial y} = \frac{\partial}{\partial y}(3y^3) = 9y^2\)

Partial derivative with respect to \(z\):

\(\frac{\partial U}{\partial z} = \frac{\partial}{\partial z}(2z) = 2\)

Substituting these partial derivatives yields the force vector:

\(\vec{F} = - (4x, 9y^2, 2)\)

To find the \(x\)-component of the force at \(P(1, 2, 3)\), substitute \(x = 1\), \(y = 2\), and \(z = 3\):

\(F_x = -4x = -4(1) = -4 \, \text{N}\)

The magnitude of the \(x\)-component of the force is:

\(|F_x| = 4 \, \text{N}\)

Therefore, the magnitude of the \(x\)-component of the force acting on the particle at point \(P(1, 2, 3)\) is 4 N.