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Question
Write down the expression for the elastic potential energy of a stretched wire.
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Solution
The work done in stretching the wire by dl,
dW = F.dl
The total work done in stretching the wire from 0 to l is
W = `int_0^"l" "F"."dl"` ..........(1)
From Young’s modulus of elasticity, force becomes,
Y = `"F"/"A" xx "L"/"l" = "YAl"/"L"` ........(2)
Substituting equation (2) in (1) we get,
W = `int_0^"l" "YAl"/"L" "dl"`
Since l is the dummy variable in the integration, we can change l to lʹ(not in limits).
Therefore W = `int_0^"l" "YAlʹ"/"L" "dlʹ" = "YA"/"L"["lʹ"^2/2]_0^"l" = "YA"/"L" "l"^2/2 = 1/2["YAl"/"L"] "l" = 1/2 "Fl"`
W = `1/2` Fl
This work done is known as the elastic potential energy of a stretched wire.
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