A steel wire of 4.0 m in length is stretched through 2.00 mm. The cross-sectional area of the wire is `2.0mm^(2)` if young's modulus of steel is `2.0xx10^(11)N//m^(2)` find (i) the energy density of wire (ii) the elastic potential energy stored in the wire.

Explanation

Here , l=4.0 m , `Deltal=2xx10^(-3)` m , `a=2.0xx10^(-6) m^2 , Y=2.0xx10^11 N//m^2`
(i) The energy density of stretched wire
`u=1/2` x stress x stain
`=1/2 x Y x "(strain)"^2`
`=1/2xx2.0xx10^11 xx(2xx10^(-3))//4)^2`
`=0.25xx10^5 = 2.5xx10^4 J//m^3`
(ii) Elastic potential energy = energy density x volume
`=2.5xx10^4xx(2.0xx10^(-6))xx4.0J = 20xx10^(-2)` = 0.20 J