Question

One end of a slack wire (Young’s modulus $Y,$ length $L$ and cross-sectional area $A$) is clamped to a rigid wall and the other end to a block (mass $m$) which rests on a smooth horizontal plane. The block is set in motion with a speed $v$. What is the maximum distance the block will travel after the wire becomes taut?
$\begin{align}
  & \text{A}\text{. }v\sqrt{\dfrac{mL}{AY}} \\
 & \text{B}\text{. }v\sqrt{\dfrac{2mL}{AY}} \\
 & \text{C}\text{. }v\sqrt{\dfrac{mL}{2AY}} \\
 & \text{D}\text{. L}\sqrt{\dfrac{mv}{AY}} \\
\end{align}$

Answer 1

Hint: When the block will be set in motion at a particular velocity, it will start executing oscillatory motion. The distance moved by the block will be equal to the extension in wire. At the maximum elongation, the strain energy of wire will be equal to the kinetic energy of the object. For calculating the distance when the wire becomes taut, we will equate the strain energy of wire to the kinetic energy of the object.

Formula used:
Expression for Strain energy,
$E=\dfrac{1}{2}VY{{\varepsilon }^{2}}$
Extension in wire,
$\varepsilon =\dfrac{x}{L}$

Complete step by step answer:
Stress is defined as the force applied on an object per unit area, or in mathematical terms, it is equal to force applied divided by the area upon which the force acts.
Strain can be defined as the change in length of object per unit length of object. This strain is the linear strain.
So strain is: $\dfrac{\text{change in length}}{\text{original length}}$
When a stress is present in an object, it leads to strain, or vice versa.
We are given that one end of a slack wire is clamped to a rigid wall and the other end to a block which rests on a smooth horizontal plane.

The strain energy of an elastic wire can be defined as the energy stored in an elastic body, when a load is applied on it. It is a result of stress and strain produced in the body.
$E=\dfrac{1}{2}VY{{\varepsilon }^{2}}$
Where,
$V$ is the volume of the object, $Y$ is the Young’s Modulus of the object and $\varepsilon $ is the strain.
Now, for the given wire,

$V$ will be the product of cross-sectional area and length of wire,
So, $V=AL$
Let, $x$ be the maximum extension in wire,
So, $\varepsilon =\dfrac{x}{L}$
So,

$\begin{align}
  & E=\dfrac{1}{2}VY{{\varepsilon }^{2}} \\
 & E=\dfrac{1}{2}\left( AL \right)Y{{\left( \dfrac{x}{L} \right)}^{2}} \\
 & E=\dfrac{1}{2}ALY{{\left( \dfrac{x}{L} \right)}^{2}} \\
\end{align}$

Now, distance moved by the block will be equal to the extension in wire, so maximum distance the block travels will be the maximum extension in wire, i.e. $x$
Now, at the point of maximum extension, the block will momentarily come to rest,
Thus all the kinetic energy will be converted into strain energy at that point.
Kinetic energy of the block: $\dfrac{1}{2}m{{v}^{2}}$
So, equating kinetic energy with strain energy, we get:

\[\begin{align}
  & \dfrac{1}{2}ALY{{\left( \dfrac{x}{L} \right)}^{2}}=\dfrac{1}{2}m{{v}^{2}} \\
 & ALY{{\left( \dfrac{x}{L} \right)}^{2}}=m{{v}^{2}} \\
 & {{x}^{2}}=\dfrac{m{{v}^{2}}L}{AY} \\
 & x=\sqrt{\dfrac{m{{v}^{2}}L}{AY}} \\
 & x=v\sqrt{\dfrac{mL}{AY}} \\
\end{align}\]

The maximum distance moved by the block will be \[v\sqrt{\dfrac{mL}{AY}}\]
Hence, the correct answer is option A.

Note:
While the block performs the oscillatory motion, there comes a point where Strain energy of the wire will be equal to the kinetic energy of the block (Or we can say, according to the law of conservation of energy, the strain energy of wire will be converted to the kinetic energy of the block). This point is known as the maximum elongation point of the wire.