Chapter 14.2, Problem 18R

To determine

The buoyant force on the block and mass of the water displaced by the block.

Answer to Problem 18R

The mass of water displace by the block is $0.120\text{kg}$ and the buoyant force on the block is $1.176\text{N}$ .

Explanation of Solution

Given info:

Density of water, $\rho =1\text{g}/{\text{cm}}^3$

Volume of the water displaced by the block, $V=120{\text{cm}}^3$

Formula Used:

Write the expression to calculate the mass of water displace by the block.

  $m=\rho \cdot V$

Write the expression to calculate the buoyant force on the block.

  $F=\rho gV$

Calculation:

Consider the conversion factor to convert density from $\text{g}/{\text{cm}}^3$ to $\text{kg}/{\text{m}}^3$ .

  $1\text{g}/{\text{cm}}^3\times \frac{1\times {10}^3\text{kg}/{\text{m}}^3}{1\text{g}/{\text{cm}}^3}=1000\text{kg}/{\text{m}}^3$

Consider the conversion factor to convert volume from ${\text{cm}}^3$ to ${\text{m}}^3$ .

  $120{\text{cm}}^3\times \frac{1\times {10}^{-6}{\text{m}}^3}{1{\text{cm}}^3}=1.20\times {10}^{-4}{\text{m}}^3$

Substitute all the values in the above expression.

  $\begin{array}{c}m=\left(1000\text{kg}/{\text{m}}^3\right)\left(1.20\times {10}^{-4}{\text{m}}^3\right)\\ =0.120\text{kg}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}F=\left(1000\text{kg}/{\text{m}}^3\right)\left(9.8\text{m}/{\text{s}}^2\right)\left(1.20\times {10}^{-4}{\text{m}}^3\right)\\ =1.176\text{N}\end{array}$

Conclusion:

Thus, the mass of water displace by the block is $0.120\text{kg}$ and the buoyant force on the block is $1.176\text{N}$ .