Question

A wire of length $L$ and cross-sectional area$A$ is made of material of Young’s modulus. If the wire is stretched by an amount $x$, the work done is ………..?

Answer 1

The linearly elastic material is defined as the material that behaves elastically and also exhibits a linear relationship between stress and strains it is called linearly elastic material. Wires obey Hooke’s law, which says that the force required to stretch the spring is directly proportional to the amount of stretch. The property of the material tells us how easily it can stretch and deform. It is defined as the ratio of the tensile stress to tensile strain naming Young’s modulus.

Formula Used:
$E = \dfrac{\sigma }{\xi }$
Here$E$ is the representation of young’s modulus of the material
$\sigma$ is the tensile stress
$\xi$ is the tensile strain.

Complete step by step solution:
From the above formula,
Stress can be defined as the total amount of force acting on the material per unit area and strain is defined as the change in length to the original length.
Stress $\sigma$=$\dfrac{F}{A}$
Here the Force acting on the wire is $F$
$A$ is the area
Strain $\xi$=$\dfrac{{dl}}{l}$
Here $dl$ is the change in length
$l$ is the actual length
From the question, we have
Actual length $l$=$L$
Area $A$=$A$
Change in length $dl$=$x$,
Let the young’s modulus of the material be $Y$, substituting this in the formula we get
$Y = \dfrac{{F/A}}{{x/L}}$
Rearranging the above expression to get an expression for force,
$Y = \dfrac{F}{A} \times \dfrac{L}{x}$
$F = \dfrac{{Y \times A \times x}}{L}$
Now to calculate the work done to stretch the wire through $x$ we have a formula,
$dw = F.dx$
Here, $dw$ is the work done
$dx$ is the average extension
Therefore we can calculate the total work done as,
$\int {dw} = \int_0^x {F.dx}$
$\int {dw} = \int_0^x {\dfrac{{YAx}}{L}dx}$
$W = \dfrac{{YA{x^2}}}{{2L}}$
The above value gives us the total work done in stretching the wire to a length $x$.

Note:
The unit of Young's modulus is Pascal $\left( P \right)$ or $N/{m^2}$. As strain is a dimensionless quantity the unit of the young’s modulus will be the same as the unit of stress that is Pascal $\left( P \right)$ or $N/{m^2}$. The value of Young’s modulus for steel or glass is comparatively larger than wood or plastic. By understanding this we can claim that steel or glass is more rigid in nature than wood or plastic. The values of Young’s modulus are given below.
Steel – $200$
Glass – $65$
Wood – $13$
Plastic (Polystyrene) – $3$