A wooden block of volume 5.22 x 10-4 m3 floats in water, and a small steel objec

Question

A wooden block of volume 5.22 x 10-4 m3 floats in water, and a small steel object of mass m is placed on top of the block. when m = 0.320 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/cm3.) (a) What is the density of the wood? kg/m3 (b) What happens to the block when the steel object is replaced by an object whose mass is less than 0.320 kg? O The block will float higher in the water. O The position of the block will remain unchanged O The block will sink slightly. O 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 0.390 kg? O The block will float higher in the water. O The position of the block will remain unchanged O The block will sink slightly. O The block will sink to the bottom.

Explanation / Answer

Volume of water displaced = Volume of the wooden block

Weight of water displaced = mass of water displaced * g

= volume of water displaced * density of water * g

= 5.22 x 10^-4 * 1000 * g

= 0.522g

Therefore

buoyant force = weight of water displaced = 0.522g

This buoyant force balances the weight of the wooden block and the weight of the steel object.

So

Mg + mg = 0.522g

where M is the mass of the wooden block

or M + m = 0.522

or M = 0.522 - 0.320 = 0.202 kg

Density of wood = Mass / volume = 0.202 / (5.22 x 10^-4)

= 386.97 kg/m^3

b)

The block will float higher in water.

c)

The block will sink to the bottom.

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