Question

A uniform chain of length 2 m and mass 0.1 kg overchangs a smooth table with its two third parts lying on the table. Find the kinetic energy of the chain as it completely slips-off the table.

Answer 1

Hint: The distribution of the mass is constant. One third of the components are hanging and two thirds are on the table. The weight will be near the centre of the hanging part's mass because one third of the hanging mass is \[\dfrac{m}{3}\].

Complete step by step solution:
Since the mass of the dangling portion is currently \[\dfrac{m}{3}\], we may use the work energy theorem to find a solution. The whole length is \[l\], and the hanging portion will be \[\dfrac{l}{3}\] in length. The centre of mass of this dangling component will be at the middle point, which is at \[\dfrac{l}{6}\], because the mass distribution is uniform.
Hanging part’s potential energy is,
\[ = mgh\]
Now, let’s substitute the corresponding values in the above formula,
\[ \Rightarrow \dfrac{m}{3} \times g \times \dfrac{l}{6}\]

On simplifying we get,
\[W \Rightarrow \dfrac{{mgl}}{{18}}\]
The potential energy is this. According to the work energy theorem, the amount of work done is equal to the change in the object's potential energy. Therefore, this dangling component has no potential energy when it is raised.
Now, it is known that \[W \Rightarrow \]Potential energy.
On substituting the known values from the given question, we get
\[{\rm{W}} = \dfrac{{0.1 \times 9.81 \times 2}}{{18}}\]
Simplify the above we get,
\[ W = 0.10\]
Therefore, the kinetic energy of the chain as it completely slips-off the table is \[0.10mgl\].

Note: Students should be careful while applying formulas. According to the work energy theorem, the amount of work done is equal to the change in the object's potential energy. The mass of the rope is constant, m. The rope's centre of mass is thus located halfway along its length.