Buoyancy and Newton’s Laws: Worked Solutions

Try the questions on the lesson page first. Then compare both the answer and the reasoning. The worked steps here are original.

Question 1: A light block is fully submerged and initially accelerates upward after release. Which statement is correct?

A light block is fully submerged and initially accelerates upward after release. Which statement is correct?

1. Draw upward buoyancy and downward weight. Why: Both forces act on the released submerged block.
2. Use F_B − mg = ma with upward positive. Why: Negligible other forces makes this the full vertical balance.
3. Upward acceleration requires F_B > mg. Why: A positive net force changes velocity upward; equality would give zero acceleration.

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Question 2: A rigid fully submerged block is lowered deeper into the same incompressible fluid. Its buoyant force…

A rigid fully submerged block is lowered deeper into the same incompressible fluid. Its buoyant force…

1. Write F_B = ρ_fluid V_displaced g. Why: Buoyancy equals the weight of displaced fluid in this hydrostatic model.
2. Keep density and displaced volume fixed while lowering the rigid block. Why: The incompressible fluid and fully submerged rigid object do not change these quantities.
3. Buoyancy stays unchanged. Why: The top and bottom pressures both increase by the same amount, leaving their difference fixed.

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Question 3: A freely floating object has average density 600 kg/m³ in water of density 1000 kg/m³. Its submerged volume fraction is…

A freely floating object has average density 600 kg/m³ in water of density 1000 kg/m³. Its submerged volume fraction is…

1. At free floating equilibrium, set ρ_fluid V_sub g = ρ_object V_total g. Why: Buoyancy and weight balance when no other vertical forces act.
2. Divide to get V_sub/V_total = ρ_object/ρ_fluid. Why: The volume fraction is determined by the density ratio.
3. Calculate 600/1000 = 0.60. Why: Sixty percent of the volume is submerged; height fraction need not be the same for an irregular shape.

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Question 4: same buoyancy, different acceleration

Two rigid objects with equal external volume V are fully submerged in the same fluid. Object A has mass m and object B has mass 2m. Derive their initial upward accelerations when released from rest with drag neglected. Is B’s acceleration always half of A’s?

1. Both objects displace V, so both have F_B = ρVg. Why: They share external volume and are fully submerged in the same fluid.
2. Apply Newton’s law: a_A = ρVg/m − g, and a_B = ρVg/(2m) − g. Why: Mass changes both weight and inertia, so buoyancy alone cannot be treated as the net force.
3. Substitute ρVg/m = a_A + g to obtain a_B = (a_A − g)/2. Why: The extra gravity term shows why B’s acceleration is not generally half of A’s. The accelerations can even have opposite signs.

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