Homework Help: physics! Need help badly

Posted by Kristen on Wednesday, April 18, 2012 at 10:54pm.

A 4.65 kg object is released from rest while fully submerged in a liquid. The liquid displaced by the submerged object has a mass of 2.60 kg. How far and in what direction does the object move in 0.300 s, assuming that it moves freely and that the drag force on it from the liquid is negligible?

physics! Need help badly - Kristen, Wednesday, April 18, 2012 at 11:01pm
never mind I finally got it after looking at it for almost and hour.
physics! Need help badly - tek, Thursday, November 8, 2012 at 8:09pm
COOL STORY, BRO. NEVER MIND POSTING HOW YOU SOLVED IT.

Anyway, here's how you solve it:
m = mass of liquid displaced
M = mass of object
F(b) = buoyant force

What two vertical forces are acting on this object? The buoyant force pushing up and the weight of the object.

F(b) = mg (pushing up)
F(weight) = Mg (pulling down)

So our net forces can be written as:
F(net) = Mg - mg = (M - m)g

And you also know that the sum of forces equals the mass times acceleration.
F(net) = Ma

So, rewritten:
(M - m)g = Ma

Do some algebra:
(M-m)*g/M = a

And I'm sure you remember the constant acceleration
x = 1/2 * a * t^2

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