A cylindrical wooden float whose base area S and the height H drift on the water surface. Density of wood d and density of water is `rho`. What minimum work must be performed to take the float out of the water ?

Answer

`Sgd^(2)H^(2)`

Explanation

Applying Newton's law vertical direction

`mg = F_(B)`
`rArr dSH xx g = rho_(w) Sh xx g`
`rArr h = (dH)/(rho)`

Now when force F is applied, for minimum work a
= 0 (`because` for a =0, F is minimum)
`F-mg +rhoSxxg=0`
`F = mg -rhoS xx g`
`W = int Fdx = int(mg -rhoS xxg)dx`
`=mg int dx - rho Sg int xdx`
`W =mgh-(rhoSgh^(2))/(2)=(rhoSh)gh-(rhoSgh^(2))/(2)`
`=(rhoSgh^(2))/(2)=(rhoSg)/(2)((dH)/(rho))^(2)=(Sg d^(2)H^(2))/(2 rho)`