Question

A block of mass $m$ held touching the upper end of a light spring of force constant $K$ as shown in the figure. Find the maximum potential energy stored in the spring if the block is released suddenly on the spring.

• A. $\frac{m^2{g}^2}{k}$
• B. $\frac{2m^2{g}^2}{k}$
• C. $\frac{4m^2{g}^2}{k}$
• D. $\frac{m^2{g}^2}{2k}$

Answer 1

The correct option is B $\frac{2m^2{g}^2}{k}$

Since the system is conservative, the total energy $E$ of the system remains constant
i.e. $E=$ constant.
$\Rightarrow \Delta E=0$
$\Rightarrow \Delta U+\Delta K.E=0$
$\Rightarrow (\Delta U{)}_{spring}+(\Delta U{)}_{gr}+\Delta K.E=0...\left(1\right)$

where $(\Delta U{)}_{spring}=$ change in potential energy of the spring $=\frac{1}{2}k{x}^2$, since the spring is compressed through a distance $x$.

$\Delta U_{gr}=$ change in gravitational potential energy of the block falls through $x,\Delta U_{gr}=-mgx$

As the block is released from rest and comes to rest after compressing the spring through a length $x$, the change in $K.E$ is zero.
i.e. $\Delta K.E=0$

Putting the values, we get,
$\frac{1}{2}k{x}^2-mgx=0\Rightarrow x=\frac{2mg}{k}$
$\therefore$ The maximum potential energy stored in the spring is,
$=\frac{1}{2}k{\left(\frac{2mg}{k}\right)}^2$
$=\frac{2m^2{g}^2}{k}$

Hence, $\left(B\right)$ is the correct answer.

Why this Question? If no external forces act (or the work done by them is zero) and the internal forces are conservative, mechanical energy of the system remains constant. This is known as the principle of conservation of mechanical energy.