A wooden block of volume 5.27 times 10^-4 m^3 floats in water, and a small steel

Question

A wooden block of volume 5.27 times 10^-4 m^3 floats in water, and a small steel object of mass m is placed on top of the block. When m = 0.400 kg, the system is in equilibrium and the top of the wooden block is at the level of the water. (Take the density of steel to be 8 g/cm^3.) (a) What is the density of the wood? kg/m^3 (b) What happens to the block when the steel object is replaced by an object whose mass is less than 0.400 kg? The block will float higher in the water. The position of the block will remain unchanged. The block will sink slightly. The block will sink to the bottom. (c) What happens to the block when the steel object is replaced by an object whose mass is 0.430 kg? The block will float higher in the water. The position of the block will remain unchanged. The block will sink slightly. The block will sink to the bottom.

Explanation / Answer

(A) for floating,

Fnet = 0

W - Fb = 0

W = (5.27 x 10^-4 x rho_wood)g + 0.4g {weight }

Fb = (5.27 x 10^-4 x rho_water)g {buoyant force }

W - Fb= 0

(5.27 x 10^-4 x rho_wood) + 0.4 - (5.27 x 10^-4 x 1000) = 0

(5.27 x 10^-4 x rho_wood) + 0.4 - 0.527 = 0

rho_wood = 241 kg/m^3

(b) now W will be lesser so we will need less buoyancy force(Fb)

hence submerged volume will be reduced.

The block will float higher in the water.

(c) now it will sink to the bottom.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.