Question

Calculate the work done in stretching steel wire of length 2m and of cross-sectional area 0.0225$m{m^2}$, when a load of 100 N is applied slowly to its free end. (Young’s modulus of steel $= \dfrac{{20 \times {{10}^{10}}N}}{{{m^2}}}$)

Answer 1

Hint The Young’s modulus is defined as the essence of the stiffness of a material. In simple words, we can say that how it can be easily bent or stretched. With this concept we have to solve this question.

Complete step by step answer:
It is given,
 $L = 2m\;A = 0.0225m{m^2} = 0.0225 \times {10^{ - 6}}{m^3}\;\;F = 100N$
Now we have to find the value of stress as:
$Stress = \dfrac{F}{A} = \dfrac{{100}}{{0.0225 \times {{10}^{ - 6}}}} = 4.4 \times {10^9}\dfrac{N}{{{m^2}}}$
For the energy stored we can write that:
$Energy\;stored\;U = \dfrac{1}{2} \times \dfrac{{{{\left( {stress} \right)}^2}}}{Y} \times AL$
Or
$U = \dfrac{1}{2} \times \dfrac{{\left( {4.4 \times {{10}^9}} \right)}}{{20 \times {{10}^{10}}}} \times \left( {0.0225 \times {{10}^{ - 6}}} \right)\left( 2 \right) = 2.222\;J$

Hence, the answer is $2.222\;J$.

Note To measure the Young’s modulus we know that we need to have an idea about the slope of the elastic stress and the strain graph. It describes the relative stiffness of a material. It is a material property and so it is not the same in all the orientation of the material. In case of the metals and the ceramics that are isotopic the value of the Young’s modulus will always have a constant value since the mechanical properties are the same for them in all forms.