The length of a steel wire is 1.0 m and its cross-sectional area is 0.03 x 10⁻⁴ m². Calculate the work done in stretching the wire when a force of 100 N is applied within the elastic region. Young's modulus of steel is 3.0 x 10¹¹ N m⁻². (Ans : 5.6 x 10⁻³ J )



Given Data:

Length of steel wire = L =  1 m 

Area = A = 0.03 x 10⁻⁴ m²

Force = F = 100 N

Young's modulus of steel = Y = 3.0 ｘ 10¹¹ N m⁻²



To Find:


Work done = W = ?



Solution:


The formula for work done is

 Work done = W = `\frac {1}{2}` F ｘ ΔL -------(1)

But the value of ΔL is unknown. So to find its value we are using the Young Modulus Y formula

Y = `\frac {σ}{Ɛ}`

or

Y = `\frac {F/A}{{ΔL}/L}`

or

Y = `\frac {FL}{AΔL}`

or

ΔL = `\frac {FL}{AY}`

by putting values

ΔL = `\frac {100 N ｘ 1 m}{0.03 x 10⁻⁴ m² ｘ 3.0 ｘ 10¹¹ N m⁻²}`

ΔL = `\frac {100 N m}{0.09 x 10⁷ N }`

ΔL = 1111.111 x 10⁻⁷ m

or

ΔL = 1.111 x 10⁻⁴ m


Now putting the corresponding values in equation (1)

W = `\frac {1}{2}` 100 N ｘ 1.111 ｘ 10⁻⁴ m 

W = 50N ｘ 1.111 ｘ 10⁻⁴ m 

W = 55.55 x 10⁻⁴ m 

or

W = 5.555 x 10⁻⁵ m --------------------Ans.




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