Two opposite forces`F_(1) = 120 N` and `F_(2) = 80 N` act on an elastic plank of modulus of elasticity `Y= 2 x 10^(11) N//m^(2)` and length `l=1m` placed over a smooth horizontal surface. The cross-sectional area of the plank is `S = 0.5 m^(2)`. The change in length of the plank is `x xx 10^(-11)m`. Find the value of `x`.

Explanation

`dL=T/A(dx)/y`
`T=F_(1)-(F_(1)-F_(2))x/L`
`int_(0)^(DeltaL)dl=((F_(1)+F_(2))L)/(2Ay)=1xx10^(-9) m`