Question

What is the minimum area of the top surface of a slab of ice with a uniform thickness of 0.30 m floating on fresh water that will hold up a car of mass 1100 kg sitting at the center on the top surface? The density of water is 1.0 × 103 kg/m3 and density of ice is 0.92 × 103 kg/m3 .

Answer 1

The minimum area of the top surface of a slab of ice is 45.8 m²

Step-by-step explanation:

Given that,

Thickness = 0.30 m

Mass of car = 1100 kg

Density of water
`\rho_(w)=1.0*10^(3)\ kg/m^3`

Density of ice
`\rho_(i)=0.92*10^(3)\ kg/m^3`

We need to calculate the minimum area of the top surface of a slab of ice

Using Buoyant force force

Total weight of car + ice weight = weight of water displaced

`W_(c)+A* t*\rho_(i)* g=\rho* A* t* g`

Put the value into the formula

`1100* g+0.30* A*0.92*10^(3)* g=1.0*10^(3)* A*0.30* g`

`(1100+A*0.30*0.92*10^(3))g=1.0*10^(3)* A*0.30* g`

`1100+276 A=300A`

`A=(1100)/(24)`

`A=45.8\ m^2`

Hence, The minimum area of the top surface of a slab of ice is 45.8 m²