Chapter 19, Problem 71A

To determine

Work done to push a block to the bottom of the pool.

Answer to Problem 71A

  $W=10\text{kJ}$

Explanation of Solution

Given:

Density of wood, $\rho =500{\text{kg/m}}^3$

Volume of the wood block, $V=0.5{\text{m}}^3$

Pool depth, $d=4\text{m}$

Formula Used:

The work done is given as,

  $W=Fd$

Where, $F$ is net force and $d$ is displacement in the direction of force.

The weight is given as,

  $F_g=\rho Vg$

Where, $\rho$ is density, $V$ is volume and $g$ is gravitational acceleration.

Calculation:

To submerge the block extra force would require in downward direction.

Now, the net force is

  $\begin{array}{l}F=F_{\text{buoyancy}}-F_g\\ F_g=500\times 0.5\times 9.8\\ =2450\text{N}\end{array}$

For water,

  $\begin{array}{l}{F}_{\text{buoyancy}}=1000\times 0.5\times 9.8\\ =4900\text{N}\end{array}$

So, work required

  $\begin{array}{l}W=\left(4900-2450\right)\times 4\\ =10\text{kJ}\end{array}$

Conclusion:

Therefore, the work required is $W=10\text{kJ}$