Question

The potential energy function of a particle is given by $U=-(x^2+y^2+z^2)\text{J}$, where $x,y$ and $z$ are in meters. Find the force acting on the particle at point $A(1\text{m},3\text{m},5\text{m})$.

• A. $\sqrt{8}\text{N}$
• B. $\sqrt{136}\text{N}$
• C. $\sqrt{140}\text{N}$
• D. $10\sqrt{14}\text{N}$

Answer 1

The correct option is C $\sqrt{140}\text{N}$

Given: $U=-(x^2+y^2+z^2)\text{J}$
Force associated,
$F=-(\frac{dU}{\partial x}\hat{i}+\frac{dU}{\partial y}\hat{j}+\frac{dU}{\partial z}\hat{k})$
$F_x=-\frac{\partial U}{\partial x}=2x$
$F_y=-\frac{\partial U}{\partial y}=2y$
$F_z=-\frac{\partial U}{\partial z}=2z$
$\therefore \overrightarrow{F}=2x\hat{i}+2y\hat{j}+2z\hat{k}$

Now ${\overrightarrow{F}}_{(1,3,5)}=(2\hat{i}+6\hat{j}+10\hat{k})\text{N}$
Magnitude of force $F=\sqrt{2^2+6^2+{10}^2}$
$F=\sqrt{140}\text{N}$