Question

A block of mass m is suspended through a spring of spring constant k and is in equilibrium. A sharp blow gives the block an initial downward velocity v. How far below the equilibrium position, the block comes to an instantaneous rest?

• A. $h=2v\sqrt{\frac{m}{k}}$
• B. $h=v\sqrt{\frac{2m}{k}}$
• C. $h=v\sqrt{\frac{m}{2k}}$
• D. $h=v\sqrt{\frac{m}{k}}$

Answer 1

The correct option is D $h=v\sqrt{\frac{m}{k}}$


Let us consider the block + the spring + the earth as the system. The system has gravitational potential energy corresponding to the force between the block and the earth as well as the elastic potential energy corresponding to the spring-force. The total mechanical energy includes kinetic energy, gravitational potential energy and elastic potential energy.
When the block is in equilibrium, it is acted upon by two forces, (a) the force of gravity mg and (b) the tension in the spring T = kx, where x is the elongation. For equilibrium, mg = kx, so that the spring is stretched by a length $x=\frac{mg}{k}$. The potential energy of the spring in this position is
$\frac{1}{2}k{\left(\frac{mg}{k}\right)}^2=\frac{m^2{g}^2}{2k}$
Take the gravitational potential energy to be zero in this position. The total mechanical energy of the system just after the blow is
$\frac{1}{2}m{v}^2+\frac{m^2{g}^2}{2k}$
The only external force on this system is that due to the ceiling which does no work. Hence, the mechanical energy of this system remains constant. If the block descends through a height h before coming to an instantaneous rest, the elastic potential energy becomes $\frac{1}{2}k{\left(\frac{mg}{k+h}\right)}^2$ and the gravitational potential energy - mgh. The kinetic energy is zero in this state. Thus, we have
$\frac{1}{2}m{v}^2+\frac{m^2{g}^2}{2k}=\frac{1}{2}k{\left(\frac{mg}{k+h}\right)}^2-mgh$
Solving this we get,
$h=v\sqrt{\frac{m}{k}}$.