A polar bear of mass 200 kg stands on an ice floe that is a uniform 1.0 m thick. What is the minimum area of the floe that will just support the bear in saltwater of density $1030 \mathrm {~kg} / \mathrm { m } ^ { 3 } ?$ The density of ice is $980 \mathrm {~kg} / \mathrm { m } ^ { 3 }$

Question

A polar bear of mass  200 kg stands on an ice floe that is a uniform  1.0 m thick. What is the minimum area of the floe that will just support the bear in saltwater of density $1030 \mathrm {~kg} / \mathrm { m } ^ { 3 } ?$ The density of ice is $980 \mathrm {~kg} / \mathrm { m } ^ { 3 }$

Quizplus · Accepted Answer

To support the bear in saltwater, the weight of the water displaced by the submerged part of the ice floe must equal the total weight of the bear plus the ice floe. The weight of the bear is $200 \, \text{kg} \times 9.8 \, \text{m/s}^2$. The volume of ice $V$ needed to displace a weight equal to the bear plus the ice itself can be found using the principle of buoyancy, where the weight of the displaced water equals the weight of the bear plus the weight of the ice. Let $A$ be the area of the ice floe. The volume of the ice is $V = A \times 1.0 \, \text{m}$ (since the thickness is 1.0 m), and its weight is $980 \, \text{kg/m}^3 \times V$. The weight of the displaced saltwater is $1030 \, \text{kg/m}^3 \times V$. Setting the weight of the displaced water equal to the total weight gives:$$1030 \, V = 200 \times 9.8 + 980 \, V$$Solving for $V$ gives the volume of ice needed to support the bear. Since $V = A \times 1.0$, solving for $A$ gives the area of the ice floe. Simplifying and solving the equation will lead to the area required, which is $4.0 \, \text{m}^2$, making option B the correct answer.

A Polar Bear of Mass 200 Kg Stands on an Ice