Answer:
671.76 kg or 6590 N
Step-by-step explanation:
So the buoyant force generated by the floating ice is equals to the mass of water displaced by the submerged ice. We also need to account for gravity of ice. The resulting additional mass that the ice sheet can support is the difference between the mass of water displaced by ice and the mass of ice submerged totally in water.
![m = m_w - m_i](https://img.qammunity.org/2021/formulas/physics/college/ad1en8ouz3qv8dso6czs2d4v9ckskklw1s.png)
![m = V\rho_w - V\rho_i](https://img.qammunity.org/2021/formulas/physics/college/8jg89gl3bgr1nghn66h5jhpyrqw4hc7afz.png)
![m = V(\rho_w - \rho_i) = 6.22*(1025 - 917) = 671.76 kg](https://img.qammunity.org/2021/formulas/physics/college/1liasfo202x5v8pe1s61iiibbml87rp3kb.png)
So the ice piece can support an additional 671.76 kg of bear, or 671.76 * 9.81 = 6590 N