Question

A piece of wood is 0.570 m long, 0.270 m wide, and 0.072 m thick. Its density is 700 kg/m3. a)What volume of lead must be fastened underneath it to sink the wood in calm water so that its top is just even with the water level? Express your answer using two significant figures. b)What is the mass of this volume of lead? Express your answer using two significant figures.

Answer 1

volume of wood= (0.570 m)(0.270 m)( 0.072m)

Volume of wood = 0.01108 m^3

density of wood = 700 kg/m^3

Calculate the mass of wood .

Mass = volume * density

mass of wood= 700( 0.01108) = 7.756 kg

density of water = 1000 kg/m^3

density (lead) = 11340 kg/m^3

Let the volume of lead = V

For equilibrium of calm water

Gravatational force = boyant force (on wood and lead)

{ 7.756 + V( 11340 ) } 9.8 = 9.8 ( 1000) ( 0.01108 + V)

7.756+ 11340 V = 1000 V + 11.08

3.324 = 10340 V

V (lead) = 3.2 x 10^-4 m^3

Part b.

mass of the lead = volume (lead) * density of lead

= 3.215*10^-4 m^3* 11340kg/m³

= 3.645 kg

mass of lead =3.6  kg apprx