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 = 1212 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 = σƐ

or

Y = FAΔLL

or

Y = FLAΔL

or

ΔL = FLAY

by putting values

ΔL = 100Nｘ1m0.03x10⁻⁴m²ｘ3.0ｘ10¹¹Nm⁻²

ΔL = 100Nm0.09x10⁷N

ΔL = 1111.111 x 10⁻⁷ m

or

ΔL = 1.111 x 10⁻⁴ m


Now putting the corresponding values in equation (1)

W = 12 100 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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