Question

A block of mass $2kg$ is kept at origin at $t=0$ and is having velocity $4\sqrt{5} \, ms^{- 1}$ in positive x-direction. The only force acting on it is a conservative and its potential energy is defined as $U=-x^{3} \, + \, 6x^{2} \, + \, 15$ (SI units). Its velocity when its acceleration is minimum after $t=0$ is

• A. $8 \, ms^{- 1}$
• B. $4 \, ms^{- 1}$
• C. $10\sqrt{24} \, ms^{- 1}$
• D. $20 \, ms^{- 1}$

Answer 1

Answer: (A)

The expression of force can be calculated as
$F=-\frac{d U}{d x}\Rightarrow F=3x^{2} \, - \, 12x$
Now, for minimum force,
$\Rightarrow \frac{d F}{d x}=0 \, \Rightarrow 6x-12=0\Rightarrow x=2 \, m$
Therefore, the force as well as the acceleration is minimum at $x=2$
From energy conservation, $U_{i} \, + \, K_{i}=U_{f} \, + \, K_{f}$
$\Rightarrow \left[15\right]+\left[\frac{1}{2} \times 2 \times 80\right]=\left[- \left(2\right)^{3} + 6 \times \left(2\right)^{2} + 15\right]+\left[\frac{1}{2} \times 2 \times v^{2}\right]\Rightarrow 15 \, + \, 80=31+ \, v^{2}$
$\Rightarrow v^{2}=64$
$\Rightarrow v=8ms^{- 1}$