# Limitations Of Work-Energy Theorem

Consider the given system which shows a chain $$AB$$ of length $$l$$ and the end $$A$$ is held at rest.

Suppose we release the chain at time $$t=0$$. How do we find the velocity of the chain when the end $$A$$ is leaving the tube?

My approach was to use the work energy theorem. Let velocity of chain at time $$t$$ when it is leaving the tube be $$v$$. Also assume the chain has uniform mass distribution with linear mass density $$\lambda$$. $$W_{\textrm{gravity}}+W_{\textrm{normal}}=\Delta K=\frac{1}{2}(\lambda l)v^{2}$$ Work done by normal force can be taken as 0 as the point of application of force undergoes no displacement. $$W_{\textrm{gravity}}=\lambda(l-h)g\frac{h}{2}+\lambda(h)g\frac{h}{2}=\frac{\lambda ghl}{2}$$ This gives $$v=\sqrt{gh}$$ which is a contradictory result.

What is my mistake in the given argument and is there any discrepancy by writing the work-energy theorem (work done by net force as change in kinetic energy) here?

• The collision of the chain with the ground is inelastic. If it were elastic the chain would have to bounce upwards from the ground like a bouncing ball. This means energy is lost as heat so we cannot apply conservation of energy to the problem. Sep 22 '21 at 6:07
• There is a detailed discussion of this in the question Missing force in the system. Sep 22 '21 at 6:13
• @JohnRennie Is the chain really colliding with the ground? I thought the links of the chain undergo elastic collision. Keep in mind, the end B of the chain is already touching the ground at $t=0$. Sep 22 '21 at 8:44

According to me your answer is mostly correct, however, I would have considered that the chain undergoes circular motion in the very last bit of its motion. Hence,

$$\frac {mv^2}{r} = mg\cos\theta$$ (considering v as the final velocity, and $$r$$ as the radius)

Hence, $$v=\sqrt {rg}\cos\theta$$

This is however my approach and I welcome any counter-opinions!