The density of ice is 917 kg/m3, and the density of sea water is 1025 kg/m3. As shown is the figure, a polar bear climbs onto a piece of ice, which has a volume of 5.75 m3, and is floating in sea water. a) What is the mass of the piece of ice? b) What is the mass of the heaviest polar bear that the ice can support without sinking completely beneath the water?
Archimedes principle: the bouyant force is equal to the weight of the water displaced.
a. mass of ice=density*volume
b. weight of water displced: densitysea*volumeice*g
weight of bear and ice: =
massbear*g+massice*g
set the two weights equal, and solve for mass bear.
To find the answers to these questions, we can use the concept of buoyancy and Archimedes' principle.
a) The mass of the piece of ice can be found by multiplying its volume by its density. So, the mass (m1) of the ice is given by:
m1 = density of ice * volume of ice
= 917 kg/m3 * 5.75 m3
≈ 5273.75 kg
b) To find the maximum mass of the polar bear that the ice can support without sinking completely, we need to consider the buoyant force exerted by the water on the ice.
The buoyant force (Fb) is given by the weight of the water displaced by the ice, based on Archimedes' principle. The weight (F1) of the ice is given by its mass (m1) multiplied by the acceleration due to gravity (g).
F1 = m1 * g
The weight (F2) of the water displaced by the ice is given by the weight of the water with the same volume as the ice. The volume (V2) of the water displaced is equal to the volume of the ice.
F2 = density of sea water * g * V2
For equilibrium, the buoyant force (Fb) should be greater than or equal to the weight of the polar bear (F3), which is given by its mass (m3) multiplied by the acceleration due to gravity (g).
Fb ≥ F3
Fb ≥ m3 * g
Since the ice is floating, the buoyant force (Fb) is equal to the weight of the ice (F1) plus the weight of the water displaced (F2).
Fb = F1 + F2
Fb = m1 * g + density of sea water * g * V2
Setting Fb ≥ F3 and substituting the values, we can solve for the mass of the heaviest polar bear (m3).
m1 * g + density of sea water * g * V2 ≥ m3 * g
Simplifying the equation:
m1 + density of sea water * V2 ≥ m3
Substituting the given values:
5273.75 kg + 1025 kg/m3 * 5.75 m3 ≥ m3
Solving the inequality:
5273.75 kg + 5881.25 kg ≥ m3
m3 ≤ 11155 kg
Hence, the mass of the heaviest polar bear that the ice can support without sinking completely beneath the water is 11155 kg.