(a) A wire 4 m long and 0.3mm in diameter is stretched by a force of 100 N. If extension in the wire is 0.3 mm, calculate the potential energy stored in the wire.
(b) Find the work done is stretching a wire of cross-section `1mm^(2)` and length 2 m through 0.1 mm. Young's modulus for the material of wire is `2.0xx10^(11) Nm^(-2)`.

Explanation

(a) Energy stored
`U=(1)/(2)` (stress)(strain)(volume)
`U=(1)/(2)((F)/(A))((Deltal)/(l)) (Al)=(1)/(2)F.Deltal`
`=(1)/(2)(100)(0.3xx10^(-3))=0.015 J`
(b) Work done =Potential energy stored
`=(1)/(2)k(Deltal)^(2)=(1)/(2)((YA)/(l))(Deltal)^(2)" "("as "k=(YA)/(l))`
Substituting the values, we have
`W=(1)/(2)((2.0xx10^(11))(10^(-6)))/((2))(0.1xx10^(-3))^(2)=5.0xx10^(-4) J`