A steel wire of length 3.0m is stretched through 3.0 mm. the cross-sectional area of the wire is `5.0mm^(2)`. Calculate the elastic potential energy stroed in the wire in the stretched condition. Young's modulus of steel is `2.0 xx 10^(11) Nm^(-2)`.

Explanation

Here, `l=3.0m`
`Delta l = 3.0mm = 3.0 xx 10^(-3)m` ,
`A=5.0mm^(2) = 5.0 xx 10^(-6)m^(2)`,
`Y=2.0 xx 10^(11) Nm^(-2)`
Elastic potential energy stored in stretched wire is
`U=1/2 stress xx strai n xx volume`
` = 1/2 (Y xx strai n) xx strai n xx volume`
`=1.2 xx Y xx (stari n)^(2) xx volume`
`=1/2 xx Y = ((Delta l)/(l))^(2) xx A xx l`
`=1/2 xx y xx (Delta l^(2)A)/(l)`
`1/2 xx (2.0 xx 10^(11)) xx ((3.0 xx 10^(-3))^(2))/(3.0) xx (5.0 xx 10^(-5))`
`=1.5 J`.