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
To Find:
Work done = W = ?
Solution:
The formula for work done is
Work done = W = 12 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
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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