Question

The system given in figure is released from rest with the spring initially stretched by $x$. The velocity $v$ of the block after it has dropped a distance $x/2$ will be

• A. $\left(\frac{4mgx-9k{x}^2}{4m}\right)$
• B. ${\left(\frac{mgx-3k{x}^2}{m}\right)}^{1/2}$
• C. ${\left(\frac{mgx-k{x}^2}{2m}\right)}^{1/2}$
• D. zero

Answer 1

The correct option is B ${\left(\frac{mgx-3k{x}^2}{m}\right)}^{1/2}$

If $m$ drops by a distance $x/2$ spring further stretches by $x$,

By principle of conservation of mechanical energy,

$\frac{1}{2}k{x}^2=\frac{1}{2}k{\left(2x\right)}^2-\frac{mgx}{2}+\frac{1}{2}m{v}^2$

$mgx=3k{x}^2+m{v}^2$

$v={\left(\frac{mgx-3k{x}^2}{m}\right)}^{1/2}$

$\therefore \left(b\right)$