Work Power Energy Question 2
Question 2 - 2024 (29 Jan Shift 1)
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 $\mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ are in meter. The magnitude of $\mathrm{x}$ - component of force (in $\mathrm{N}$ ) acting on the particle at point $\mathrm{P}(1,2,3) \mathrm{m}$ is :
(1) 2
(2) 6
(3) 4
(4) 8
Show Answer
Answer: (3)
Solution:
Given $U=2 x^{2}+3 y^{3}+2 z$
$\mathrm{F}_{\mathrm{x}}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}=-4 \mathrm{x}$
At $x=1$ magnitude of $F_{x}$ is $4 N$
