Question

2. You have a piece of driftwood with a volume of 3.0x10 m' submerged below the surface of a swimming pool. The density of the wood is 400kg/m. a) What buoyant force does the wood experience? b) What is the acceleration of the piece of driftwood (magnitude and direction)?

Answer 1

1) Buoyant force on piece of wood is given by

$F= \rho_w Vg$

where rho is density of water= 1000kg/m3

V is volume displaced by wood

so putting all values we get,

F= 29.4 N

2) accelartion of wood is given by newton second law

$F= ma=\rho Va$

where rho is density of wood block

so,

$a=\frac{F}{\rho V}$

so,

putting values we get,

a= 24.5 m/s2

and net accelartion of wood block is

$a_{net}=a-g$

= 14.7 m/s2 in upward direction