A child’s flotation device (ρ = 250 kg / m³) is held underwater at the beach (ρ_seawater = 1025 kg / m³). 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 / s²)

A) 2.45 m / s² downwards

B) 0 m / s²

C) 29.4 m / s² upwards

D) 39.2 m / s² 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_toya

We can then substitute for the forces:

ρ_seawaterV_toyg – m_toyg = m_toya

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_toyV_toy = m_toy

ρ_seawaterV_toyg – ρ_toyV_toyg = ρ_toyV_toya

Factor out the left side of the equation:

(ρ_seawater – ρ_toy)V_toyg = ρ_toyV_toya

(ρ_seawater – ρ_toy)g = ρ_toya

(1000 – 250)(10) = (250)a

(750)(10) = (250)a

(3)(10) = a

30 = a

The closest answer choice is (C).

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