A steel wire of length 2.0 m is stretched through 2.0 mm. The cross sectional area of the wire is 4.0 mm^2. Calculate the elastic potential energy stored in the wire in the stretched condition. Young modulus of steel `=2.0x10^11Nm^-2`

Explanation

The strain in the wire `(/_\l)/l=(2.0mm)/(2.0m)=10^-3`
The stress in the wire `=YxxStrain`
`=2.0xx10^1Nm^-2xx10^-3=2.0xx10^8Nm^-2`
The volume of the wire `=(4xx10^-6m^2)xx(2.0m)`
`=8.0xx106-6m^3`
the elastic potential energy stored
`=1/2xxstressxxstrainxxvolume`
`=1/2xx2.0xx10^8Nm^-2xx8.0xx10^-6m^3`
`=0.8J`