A slab of ice floats on a freshwater lake. What minimum volume must the slab have for a 65.0-kg woman to be able to stand on it without getting her feet wet?

Given data can be listed below as,

• Mass of the woman$m=65\mathrm{kg}$
• The slab is floating on the freshwater.

Applying Newton’s second law, the buoyancy force exerted by the water is upward and given by water Gravity pulls more on the heavier substance when a lighter object is put in a fluid. When an object's weight exceeds that of the fluid it displaces, gravity pulls harder on it, causing it to sink.

The buoyant force can be expressed as

$B=\rho Vg$ …i)

Here, V is the volume of ice,$\rho$ is the density of water, and g is the acceleration due to gravity.

The net force is given by,

$\underset{}{\sum F=ma}$

Here, m is mass, and ais acceleration.

Total force can be given by,

$B=mg=0$

By equation i),

localid="1657176820345" $$\begin{gathered} {\rho }_w{V}_{i}g=\left(65.0\mathrm{kg}+m_i\right)g \\ {\rho }_w{V}_{i}g=\left(65.0\mathrm{kg}+{\rho }_i{V}_i\right)g \\ \left({\rho }_w-{\rho }_i\right)V_i=65.0\mathrm{kg} \end{gathered}$$

Substitute all values in the above,

$$\begin{gathered} V_i=\frac{65.0\mathrm{kg}}{\left({\rho }_w-{\rho }_i\right)} \\ =\frac{65.0\mathrm{kg}}{\left(1000\mathrm{kg}/{\mathrm{m}}^3-920\mathrm{kg}/{\mathrm{m}}^3\right)}\left(\frac{\mathrm{kg}}{\mathrm{kg}/{\mathrm{m}}^3}\right) \\ =0.81{\mathrm{m}}^3 \end{gathered}$$

Thus, the minimum volume of the slab is$0.81{\mathrm{m}}^3$.