Calculate the increase in energy of a gold bar of length 0.15m, and area of cross section `15xx10^(-4)m^(2)` when compressed with a load of 10 kg weight along its length.
Given `Y=71xx10^(9)Nm^(-)` and `g=9.8ms^(-2)`

Explanation

Given `Y=7.1xx10^(10)Nm^(-2)`
`g= 9.8 ms^(-2), l=0.15m, A=15xx10^(-4)m^(2), m=10 kg`
We know that `Y=(mgL)/((Deltal)A)`
i.e. `Delta l=(mgL)/(AY)=(10xx9.8x0.15)/(15xx10^(-4)xx7.1xx10^(10))=14.7/106.5xx10^(-6)`
`Deltal=0.138xx10^(-6)m`
Increase in elastic P.E. `=1/2("Force")(Deltal)=1/2(mg)(Deltal)`
`=1/2xx10xx9.8xx0.138xx10-6`
i.e. P.E `=6.763xx10^(-6)J`