Bouyency-force problem, collage dynamics problem, Help!?

Question

the buoyancy force acting on a sphere of mass 2kg immersed underwater is 10N.

the sphere is released from a depth of 5m. With what velocity does the sphere rise to the surface of the water?

can you work through the answer, i really need to understand this problem..

Answer 1

It's been half a century since I've tried problems like this...

If I get it wrong, it's cause I'm doing it from memory and that's a long, long time ago.

1.  Newton says F = ma

2.  m = 2 kg

3.  F = -9.6 N (subtract gravitational force (2 kg x 9.8 N/kg) from bouyancy force) - the result is negative, the ball will sink!

4.  If the bouyancy force is net of gravitational force, then F = 10 N.  

5.  F=ma,  a = 2 / F

From here it is a normal acceleration problem.  Do you care that "rise to the surface of the water" may be ill-defined, since the top of the sphere comes out of the water first, and as the sphere comes to the surface the bouyancy force decreases?

Answer 2

I assume that the buoyancy force you cited is a net force--otherwise the ball will sink, since it weighs twice that.

So you have contant acceleration:

a = x'' = F/m

This is just like dropping something, but it goes up:

I'll call down (depth) positive, so the solution to your equation of motion is:

x(t) = x0 - 1/2 at^2

v(t) = -at

The ball's speed when it breaks the surface is:

vf = sqrt (2 a x0)

= sqrt (2 F x0 / m)

Now once the ball hits the surface, the dynamics gets somewhat interesting--but you can save that for a little later in the course.