Question

Chain of length $1m$ is hanging as shown in the figure. The mass of the chain is $3kg$. It is released and allowed to move, find kinetic energy (approximate) of the chain when it is completely in the air.

• A. $56J$
• B. $90J$
• C. $100J$
• D. $121J$

Answer 1

The correct option is A

$56J$

Step 1, Given Data,

Total length of chain $l=4m$

Length of the hanging chain $l_1=1m$

Length of the chain on table $l_2=3m$

Mass of the chain $m=3kg$

We have to calculate the kinetic energy of the hanging chain.

Step 2. Calculate the initial Potential Energy and final Potential Energy,

Mass of hanging chain $=\frac{1}{4}\left(3kg\right)$

(Note: Since the chain is at rest hence the initial potential energy will be the same only for the potential energy of the hanging chain and the final potential energy will be equal to the potential energy of the whole chain. )

Initial Potential Energy $=$Potential Energy of hanging chain

$$\begin{gathered} U_1=-{\int }_0^{l_1}\left(\frac{m}{l}\right)gxdxJ \\ \Rightarrow U_1=-\left(\frac{m}{l}\right)g{\int }_0^{1}xdxJ \\ \Rightarrow U_1=-\left(\frac{m}{l}\right)g{\left[\frac{x^2}{2}\right]}_0^{1}J \\ \Rightarrow U_1=-\left(\frac{3}{4}\times 10\right)\left(\frac{1}{2}\right)J \\ \Rightarrow U_1=-\frac{30}{8}J \end{gathered}$$

Final Potential Energy $=$Potential Energy of the whole chain

$$\begin{gathered} U_2=-{\int }_0^l\left(\frac{m}{l}\right)gxdxJ \\ \Rightarrow U_2=-\left(\frac{m}{l}\right)g{\int }_0^{4}xdxJ \\ \Rightarrow U_2=-\left(\frac{m}{l}\right)g{\left[\frac{x^2}{2}\right]}_0^{4}J \\ \Rightarrow U_2=-\left(\frac{3}{4}\times 10\right)\left(8\right)J \\ \Rightarrow U_2=-60J \end{gathered}$$

Step 3. Calculate the Kinetic Energy of the hanging chain,

(Note: Since we know that energy is conserved hence if kinetic energy will increase in an object then its potential energy will decrease.)

Gain of Kinetic Energy $=$Loss of Potential Energy

$K.E.=$Initial Potential Energy $-$Final Potential Energy

$$\begin{gathered} \Rightarrow K.E.=-\left(\frac{30}{8}\right)-\left(-60\right)J \\ \Rightarrow K.E.=56.25J \\ \Rightarrow K.E.\approx 56J \end{gathered}$$

Hence, option A is correct.