Question

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

• A. $1$
• B. $2$
• C. $1.5$
• D. $3$

Answer 1

The correct option is A $1$

We know that total energy is conserved.
$E_i=E_f\text{here,}{E}_i=\text{initial potential energy+initial kinetic energy}{E}_f=\text{final potential energy+final kinetic energy}$
When the whole chain is on the inclined plane initially,
its Potential energy is $mg\frac{L}{2}sin{30}^{\circ }$ and kinetic energy is zero
$\therefore E_i=mg\frac{L}{2}sin{30}^{\circ }$
When half of the chain is off the inclined plane,
the potential energy of the part that is on the inclined plane $=\frac{m}{2}g\frac{L}{4}sin{30}^{\circ }$
and for the part that has fallen $=-\frac{m}{2}g\frac{L}{4}$
If the velocity of the chain is $v$ then its kinetic energy is $\frac{1}{2}m{v}^2$
$E_f=\frac{m}{2}g\frac{L}{4}sin{30}^{\circ }+-\frac{m}{2}g\frac{L}{4}+\frac{1}{2}m{v}^2$

Using conservation of energy,
$mg\frac{L}{2}sin{30}^{\circ }=\frac{m}{2}g\frac{L}{4}sin{30}^{\circ }-\frac{m}{2}g\frac{L}{4}+\frac{1}{2}m{v}^2$

Solving we get, $v=\sqrt{\frac{5gL}{8}}$

Substituting $g$ and $L=\frac{4}{25}$ we get,
$v=1m/s$