Question

A steel wire of length 2m is hanging from a rigid horizontal support. How much energy is stored in it when a load of 5 kg is suspended on it? If the load is increased to 10 kg then by how much energy will increase? ($Y=2\times {{10}^{11}}N/{{m}^{2}}$, area of cross section $={{10}^{-6}}{{m}^{2}}$).

Answer 1

Express the potential energy of a stretched string in terms of the stress and strain. Find the stress from the given values and find strain from the obtained stress. Now find the potential energy for the two given masses and subtract them to find the required solution.

Complete answer:
When we stretch a wire or a string, we need to do work. This work done is stored as the potential energy of the wire or the string.

We can mathematically define the potential energy per unit length stored in the string in terms of the stress and strain of the wire as,
$U=\dfrac{1}{2}\times \text{stress}\times \text{strain}$
Where, U is the potential energy of the wire.
We are given that,
The length of the wire is $l=2m$
Initial load applied on the wire ${{m}_{i}}=5kg$
Final load applied on the wire ${{m}_{f}}=10kg$
The cross-section of the wire is, $A={{10}^{-6}}{{m}^{2}}$
Young’s modulus of the wire, $Y=~2\times {{10}^{11}}N/{{m}^{2}}$
Now, stress of the wire can be defined as the force or weight per unit area of the wire.
Again, strain can be expressed in terms of the stress and the young’s modulus as,
$\text{strain}=\dfrac{\text{stress}}{Y}$
When we apply the 5kg load, the potential energy per unit length of the wire will be,
${{U}_{i}}=\dfrac{1}{2}\times \dfrac{5g}{{{10}^{-6}}}\times \dfrac{\dfrac{5g}{{{10}^{-6}}}}{2\times {{10}^{11}}}=\dfrac{250g}{4}$
So, the potential energy stored in the wire is $={{U}_{i}}\times l=\dfrac{250g}{4}\times 2=125g$
When we apply 10 kg load on the wire, the potential energy per unit length of the wire will be,
${{U}_{f}}=\dfrac{1}{2}\times \dfrac{10g}{{{10}^{-6}}}\times \dfrac{\dfrac{10g}{{{10}^{-6}}}}{2\times {{10}^{11}}}=\dfrac{1000g}{4}=250g$
So, the potential energy stored in the wire is $={{U}_{f}}\times l=250g\times 2=500g$
So, the increase in potential energy in the wire will be,
$\begin{aligned} & E=500g-125g \\\ & E=375g \\\ & E=375\times 9.8J \\\ & E=3675J \\\ \end{aligned}$
Increase in energy stored in the wire will be 3675J.

Note:
The potential energy of a stretched string can be defined in terms of the displacement of the string from the equilibrium position also. We can mathematically express it as,
$E=\dfrac{1}{2}k{{x}^{2}}$
Where, k is the spring constant and x is the displacement from the equilibrium position.