Question 75: Work, Energy and Power & Circular Motion

Question

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 :

Accepted Answer

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} = -
abla U $
where $ 
abla U $ (the gradient of $ U $) is a vector with components given by the partial derivatives of $ U $ with respect to $ x $, $ y $, and $ z $:
$ 
abla U = \left( \frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}, \frac{\partial U}{\partial z} ight) $
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 \, 	ext{N} $
Thus, the correct answer is
Option A: 4
.

Answer

4

Answer

2

Answer

8

Answer

6

Work, Energy and Power & Circular Motion

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 :

Choose the correct answer: