Answer to Question #168056 in Physics for Tithi

Question

Find the work done in stretching a wire of 1 sq. mm cross-section and 2 m long through 0.1 mm. Given that the Young’s modulus of the material of the wire is 2.0 × 1011 N/m2. 

Answer 1

The work is:

$$W = \dfrac{1}{2}kx^2$$

The spring constant $k$ of the wire can be found from the Hook's law:

$$F = kx$$

where $F$ is the elastic force, and $x = 0.1mm = 10^{-4}m$ is the elongation of the wire.

On the other hand, the Hook's law states:

$$\dfrac{F}{A} = E\dfrac{x}{L}$$

where $A = 1mm^2 = 10^{-6}m^2$ is the cross-section, $L = 2m$ is the length of the wire, and $E = 2\times 10^{11}N/m^2$ is the Young’s modulus of the material of the wire.

Equating the force from these two equations, obtain:

$$kx = \dfrac{EAx}{L}\\ k = \dfrac{EA}{L}$$

Substituting this expression into the formula for the work, obtain:

$$W = \dfrac{EAx^2}{2L}\\ W = \dfrac{2\times 10^{11}\cdot 10^{-6}\cdot (10^{-4})^2}{2\cdot 2} = 5\times 10^{-4}J$$

Answer. $5\times 10^{-4}J$.