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Zorluk: KolayFluids at Rest, Archimedes' Principle and Viscosity

A solid block of wood has a mass of 0.60 kg0.60\text{ kg}. When placed in a vessel of water, it floats freely on the surface. What is the magnitude of the upthrust exerted by the water on the block? (Take g=10 m/s2g = 10\text{ m/s}^2)

  1. 6.0 N6.0\text{ N}Cevap
  2. B
    0.60 N0.60\text{ N}
  3. C
    0.0 N0.0\text{ N}
  4. D
    60 N60\text{ N}

Cevap

The upthrust exerted by the water on the floating block is 6.0 N6.0\text{ N}.
According to the Law of Flotation, a body floating freely in a fluid displaces a weight of fluid equal to its own weight. Therefore, the upward force (upthrust) exerted by the fluid is equal to the weight of the block: U=mg=0.60 kg×10 m/s2=6.0 NU = mg = 0.60\text{ kg} \times 10\text{ m/s}^2 = 6.0\text{ N}.

Adım Adım Çözüm

1
Determine the weight of the floating block in air.
W=mg=0.60 kg×10 m/s2=6.0 NW = mg = 0.60\text{ kg} \times 10\text{ m/s}^2 = 6.0\text{ N}
Weight is the force of gravity acting on the block's mass.
2
Apply the Law of Flotation to find the upthrust.
Upthrust U=W=6.0 NU = W = 6.0\text{ N}
A freely floating body displaces a volume of fluid whose weight is equal to the total weight of the body.

Anahtar Kavram

Law of Flotation and Archimedes' Principle
Tahmini Süre:45s
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