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September 3, 2014

September 3, 2014

Posted by **zoe** on Saturday, December 1, 2012 at 3:47pm.

- physics -
**Damon**, Saturday, December 1, 2012 at 8:19pmBuoyancy = 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)

- physics -
**zoe**, Sunday, December 2, 2012 at 2:03pmwhy is work done during the rise= mgh directly?

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