Question

A particle of mass $100 \, gm$ moves in a potential well given by $U=8x^{2}-4x+400 \, joule$ . Find its acceleration at a distance of $25 \, cm$ from equilibrium in the positive direction

• A. $4 \, m \, s^{- 1}$
• B. $40 \, m \, s^{- 1}$
• C. $-40 \, m \, s^{- 1}$
• D. $-4 \, m \, s^{- 1}$

Answer 1

Answer: (C)

We know that $F=-\frac{d U}{d x}$
$F=-\frac{d}{d x}\left(8 x^{2} - 4 x + 400\right)$
$F=-\left(16 x - 4\right)$
$F=4-16x$
At equilibrium $4-16x_{0}=0$ and
$x_{0}=\frac{1}{4}m=25cm$
At the given position
$x=\left(25 + 25\right)=50 \, cm$
$\Rightarrow $ $F=4-16\times 0.5=-4N$
or $a=\frac{- 4}{0.1}=-40ms^{- 2}$