Question

The potential energy of a particle is given by formula $U=100-5x+100x^2,U$ and $x$ are in SI units. If mass of the particle is $0.1\text{kg}$, then magnitude of it's acceleration

• A. At $0.05\text{m}$ from the origin is $50{\text{m/s}}^2$
• B. At $0.05\text{m}$ from the mean position is $100{\text{m/s}}^2$
• C. At $0.05\text{m}$ from the origin is $150{\text{m/s}}^2$
• D. At $0.05\text{m}$ from the mean position is $200{\text{m/s}}^2$

Answer 1

Answer: (A), (B), (C)

At $0.05\text{m}$ from the origin is $150{\text{m/s}}^2$

$U=100-5x+100x^2$
$\therefore F_x=-\frac{\partial U}{\partial x}=-\frac{\partial }{\partial x}(100-5x+100x^2)$
$=-[-5+200x]$
$=5-200x$
$\therefore a_x=\frac{F_x}{0.1}=\frac{5-200x}{0.1}$
$=50-2000x$
(i) At $0.05\text{m}$ from origin
$x=+0.05\text{m}$ (first point)
$a_x=50-2000\left(0.05\right)$
$=50-100$
$=-50{\text{m/s}}^2$
$=50{\text{m/s}}^2$ towards -ve $x$
$\therefore$ Correct option is (a)
(ii) At $0.05\text{m}$ from origin
$x=-0.05\text{m}$ (second point)
$a_x=50-2000(-0.05)$
$=50+100$
$=150{\text{m/s}}^2$
$\therefore$ Correct option is (c)
Mean Position
$a=0$ at $x=\frac{50}{2000}\text{m}=0.025\text{m}$
(iii) For point $0.05\text{m}$ from mean position
$x=(0.05+0.025)\text{m}$
$=0.075\text{m}$
$\therefore a_x(\text{at}x=0.075\text{m})=50-2000\left(0.075\right)\text{m}$
$=50-150=-100{\text{m/s}}^2$
$=100{\text{m/s}}^2$ towards -ve $x-$ axis.
$\therefore$ Correct option is (b)
(iv) For second point $0.05\text{m}$ from mean position
$x=0.025\text{m}-0.05\text{m}$
$=-0.025\text{m}$
$\therefore a_x=50-2000(-0.025)$
$=100{\text{m/s}}^{-2}$
$\therefore$ Option (d) is incorrect.