- #1

Mosaness

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**1. A slab of ice floats on a freshwater lake. What minimum volume must the slab have for a 45.0 kg woman to be able to stand on it without getting her feet wet?**

**2. Archimedes Principal as well as Newton's Second Law will be applied**

## The Attempt at a Solution

I know that the volume of water that is being displaced will equal the combined weight of the woman and the slab that is pushing down on the water.

This gives the equation:

ρVg = Weight

_{woman}+ Weight

_{ice}

***ρVg is the same as [itex]\frac{m}{V}[/itex]***

Applying this gives:

ρVg = g(45.0-kg + m

_{ice})

ρV = (45.0-kg + m

_{ice})

I know that the density for the freshwater lake will simply be 1.00 x 10

^{3}kgm

^{-3}and the density for ice will be 0.92 x 10

^{3}kgm

^{-3}.

Now, this is where I am stuck:

V = [itex]\frac{(45.0-kg + m

_{ice})}{ρ

_{water}}[/itex]

I am leaning towards saying that V should be the volume that has been displaced by the combined weight of the woman and the ice. And if I can figure out how to find that V, then I can plug it into the entire equation to give me the mass of the ice. Using the new found mass and the density of ice, I can find out the volume of the slab of ice.