(a) a spherically shaped buoy is made from wood, which has a specific gravity of 0.6. find the acceleration of the buoy when released from rest at the bottom of a freshwater lake. (b) if the buoy starts out 10 m below the surface of the water, determine the height above the water that it will rise after it shoots out of the water.

Answer 1

Buoyancy = 1000 kg/m^3 * 9.81 * vol
weight = 600 kg/m^3 * 9.81 * vol

net force up = 400 * 9.81 * vol

F = m a

400 * 9.81 * vol = 600 * vol * a

a = (2/3) * 9.81
--------------------------------
work done in rise to surface
= F * 10 = 4000*9.81*vol
= Kinetic energy on exit from water
= (1/2) m v^2
= .5 * 600 * vol * v^2
so
300 v^2 = 4000* 9.81
v^2 = 130.8
v = 11.4 m/s initial speed up

Ke = Pe at stop = m g h
but oh. I did not have to do that
work done during rise = m g h directly
4000*9.81*vol = 600*vol*9.81 * h

h = (40/6) = 6 3/3 m

note g did not matter in part b

NOte also that this problem ignores the force required to accelerate the water around the sphere. This "added mass" is actually comparable to the mass of water displaced by the sphere. I bet the text writer did not even know that (hydrodymanics, not usually covered in school or college physics but very important if you are designing fishing gear)

Answer 2

why is work done during the rise= mgh directly?