Question

The chain of length L is released from rest on a smooth fixed incline with x = 0 as shown in the figure. Velocity $v$ of the chain when a half of the length has fallen is (Take $L=\frac{4}{25}m)$ in m/s. (Neglect edge effect of inclined).

Answer 1

We know that total energy is conserved
${\epsilon }_i={\epsilon }_f{\epsilon }_i=\text{initial potential energy+initial kinetic energy}{\epsilon }_f=\text{final potential energy+final kinetic energy}mg\frac{L}{2}sin\theta +0=m_{1}g\frac{(L-x)}{2}sin\theta -m_{2}g\frac{x}{2}+\frac{1}{2}m{v}^2\text{Now if the mass is considered as uniformly distributed then}{m}_1=\frac{m}{L}(L-x)m_2=\frac{m}{L}xmg\frac{L}{2}sin\theta +0=\frac{m}{L}(L-x)g\frac{(L-x)}{2}sin\theta -\frac{m}{L}xg\frac{x}{2}+\frac{1}{2}m{v}^{2}gLsin\theta =\frac{g}{L}(L^2+x^2-2Lx)sin\theta -g\frac{x^2}{L}+v^2\text{As it is given half of the length is left so}x=\frac{L}{2}v=\sqrt{\frac{5}{8}gl}$
Taking $g=10m/s^2$
Putting value of $L=\frac{4}{25}$

$v=1m/s$