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?
3 answers
never mind I finally got it after looking at it for almost and hour.
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
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
How did you get the answer?
7 answers