Question

Calculate the energy stored in the stretched wire which will produce an extention of 0.30mm in a steel wire with a length of 4.0m and a cross sectional area of 2.0 x10^(-6) m^2 young modulus of steel is 2.1 x10^11Pa

Answer 1

Given data:

* The extension of the wire is 0.3 mm.

* The length of the wire is 4 m.

* The cross sectional area of the wire is,

$A=2.0\times10^{-6}m^2$

* The young modulus of the steel is,

$Y=2.1\times10^{11}\text{ Pa}$

Solution:

The young modulus of the steel wire in terms of the force acting on the wire is,

$Y=\frac{F\times l}{A\times dl}$

where A is the area of cross section, dl is the extension, l is the length of the wire, and F is the force acting on the wire,

Substituting the known values,

$\begin{gathered} 2.1\times10^{11}^{}=\frac{F\times4}{2.0\times10^{-6}\times0.3\times10^{-3}} \\ 2.1\times10^{11}=6.67\times10^9\times F \\ F=\frac{2.1\times10^{11}}{6.67\times10^9} \\ F=0.315\times10^2 \\ F=31.5\text{ N} \end{gathered}$

The energy stored in the stretched wire is,

$\begin{gathered} U=\frac{1}{2}\times F\times dl \\ U=\frac{1}{2}\times31.5\times0.3\times10^{-3} \\ U=4.725\times10^{-3}\text{ J} \end{gathered}$

Thus, the energy stored in the stretched wire is,

$4.725\times10^{-3}\text{ J}$