Fluids and Newton's Laws (Buoyancy)

AP Physics 1· difficulty 3/5

A 5.0 kg5.0~\text{kg} object of volume 1.0×103 m31.0 \times 10^{-3}~\text{m}^3 is fully submerged in water. The net force on it (using g10 m/s2g \approx 10~\text{m/s}^2) is

  • A

    10 N10~\text{N} down

  • B

    60 N60~\text{N} down

  • C

    00

  • D

    40 N40~\text{N} down

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Explanation

W=mg=50 NW = m g = 50~\text{N}. FB=ρwgV=10 NF_B = \rho_w g V = 10~\text{N}. Net 5010=40 N50 - 10 = 40~\text{N} down (object sinks unless held).

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