MCAT Physics Question — Flotation Device
 by
 Dec 02, 2014
 MCAT Blog, MCAT Physics, MCAT Question of the Day
 Reviewed By: Liz Flagge
A child’s flotation device (ρ = 250 kg / m^{3}) is held underwater at the beach (ρ_{seawater} = 1025 kg / m^{3}). The toy is spherical and has a radius of one meter. When the toy is released, what is the magnitude and direction of the toy’s acceleration? (assume g = 9.8 m / s^{2})
A) 2.45 m / s^{2} downwards
B) 0 m / s^{2}
C) 29.4 m / s^{2} upwards
D) 39.2 m / s^{2} upwards
Explanation
This problem asks us to solve for the net acceleration on a submerged object that is less dense than seawater. Before doing any calculations, we can see that the object’s density is approximately one fourth that of seawater. As a consequence, the object will float in seawater. Thus, when released, it will accelerate towards the water’s surface, where it will then float.
That lets us immediately eliminate choices (A) and (B). If you’re pressed for time, or are unsure how to proceed, then you can make an educated guess and move on.
To solve the problem, use Newton’s 2^{nd} law:
F_{net} = ma
The net force will be the buoyant force minus the force of gravity:
F_{b} – F_{g} = m_{toy}a
We can then substitute for the forces:
ρ_{seawater}V_{toy}g – m_{toy}g = m_{toy}a
Because this is a buoyancy problem it’s often easier to work in terms of density than mass. We can substitute ρV for m in our equation:
ρ_{toy} = m_{toy} / V_{toy}
ρ_{toy}V_{toy} = m_{toy}
ρ_{seawater}V_{toy}g – ρ_{toy}V_{toy}g = ρ_{toy}V_{toy}a
Factor out the left side of the equation:
(ρ_{seawater }– ρ_{toy})V_{toy}g = ρ_{toy}V_{toy}a
(ρ_{seawater }– ρ_{toy})g = ρ_{toy}a
(1000 – 250)(10) = (250)a
(750)(10) = (250)a
(3)(10) = a
30 = a
The closest answer choice is (C).
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