Question

A uniform elastic cord of force constant $k_0$, length $l_0$ and mass $m_0$ is lying on the ground in a heap. One end of the rope is gradually pulled upwards till the entire rope becomes vertical with its lower end touching the ground. Find the work done by the agency pulling the rope. Neglect height of the heap as compared to the length of the cord.

Answer 1

$\frac{m_0 g l_0}{2} + \frac{m_0^2 g^2}{3 k_0}$

Answer 2

The work done by the agency is the sum of the change in gravitational potential energy and the change in elastic potential energy of the cord.

1. Initial state: Cord on the ground. $PE_{gravity, initial} = 0$, $U_{elastic, initial} = 0$.
2. Final state: Cord vertical, stretched by its own weight.
   • Elastic Potential Energy ($U_{elastic, final}$):
     Consider a segment of natural length $dy_0$ at original distance $y_0$ from the bottom. Tension $T(y_0) = \frac{m_0 g}{l_0} y_0$.
     Force constant of segment $k_{dy_0} = k_0 \frac{l_0}{dy_0}$.
     Extension $d(\Delta l) = \frac{T(y_0)}{k_{dy_0}} = \frac{m_0 g y_0 dy_0}{k_0 l_0^2}$.
     Energy in segment $dU_{elastic} = \frac{1}{2} k_{dy_0} (d(\Delta l))^2 = \frac{m_0^2 g^2 y_0^2}{2 k_0 l_0^3} dy_0$.
     Total elastic energy $U_{elastic, final} = \int_0^{l_0} \frac{m_0^2 g^2 y_0^2}{2 k_0 l_0^3} dy_0 = \frac{m_0^2 g^2}{6 k_0}$.
   • Gravitational Potential Energy ($PE_{gravity, final}$):
     Height of center of mass $\bar{y} = \frac{\int y dm}{\int dm}$.
     Position of a segment $y = y_0 + \int_0^{y_0} d(\Delta l) = y_0 + \frac{m_0 g y_0^2}{2 k_0 l_0^2}$.
     $dm = \frac{m_0}{l_0} dy_0$.
     $\bar{y} = \frac{1}{l_0} \int_0^{l_0} \left(y_0 + \frac{m_0 g y_0^2}{2 k_0 l_0^2}\right) dy_0 = \frac{l_0}{2} + \frac{m_0 g}{6 k_0}$.
     Total gravitational energy $PE_{gravity, final} = m_0 g \bar{y} = \frac{m_0 g l_0}{2} + \frac{m_0^2 g^2}{6 k_0}$.
3. Total Work Done:
   $W = (PE_{gravity, final} - PE_{gravity, initial}) + (U_{elastic, final} - U_{elastic, initial})$
   $W = \left(\frac{m_0 g l_0}{2} + \frac{m_0^2 g^2}{6 k_0}\right) + \left(\frac{m_0^2 g^2}{6 k_0}\right)$
   $W = \frac{m_0 g l_0}{2} + \frac{m_0^2 g^2}{3 k_0}$.