CHECK COMPLETE QUESTION
The cube of wood having an edge dimension of 19.7cm and a density of 647kg/m3 floats on water
(a) What is the distance from the horizontal top surface of the cube to the water level answer in cm
(b) What mass of lead should be placed on the cube so that the top of the cube will be just level with the water surface answer in kg
Answer:
a)6.29cm
b)2.78 kg
Explanation:
Given:
Let us calculate the volume first, we were given dimension as 19.7cm=0.197m
Volume is (0.197 meters)³ = 0.00764m³
Then we can calculate the mass as;
Given mass is 647 kg/m³ x 0.00774m³ = 4.947kg
The weight = mass × acceleration due to gravity
weight = 4.947 x 9.8 N/kg = 48.44N
By Floating we can say the the buoyancy force has to equal the weight (48.44 N) which has
which is equal to the weight of volume of the displaced water. Or the mass, the calculation is the same.
We know that density of fresh water at 20ºC is 998 kg/m³
Then we can calculate the volume of displaced water as
4.947 kg / 998 kg/m³ = 0.00496 m³
We know that the displaced water has a shape of a rectangular solid with 0.197 meters on the two horizontal dimensions, and h as the height to the surface then
V = 0.197²h = 0.00496
0.00496= 0.197²h
h = 0.1278 meters or 12.78 cm
Then the the distance exposed, would be 19.7–12.78 = 6.29 cm
b) ifthe cube is fully submerged, the volume of the displaced water is 0.00774m³
mass of displaced water is 0.00774m³ x 998 kg/m³ = 7.724 kg
Added mass is the mass of the displaced water – mass of block
= 7.724–4.947 = 2.78 kg
