Question

A plastic ball has radius 12.0 cm and floats in water with 24.0% of its volume submerged. (a) What force must you apply to the ball to hold it at rest totally below the surface of the water? (b) If you let go of the ball, what is its acceleration the instant you release it?

Short answer

Answer

1. The force of the surface of water is, 53.96 N.
2. The acceleration is, 31.06 m/s² .

Step-by-step solution

Step-by-Step Solution Step 1: Identification of the given data

The given data can be listed below as,

• The radius of ball is, r = 12 cm .

Determination the force.

Part (a)

The relation of volume and radius of a sphere is expressed as,

$v=\frac{4}{3}\pi r^3$

Here v is the volume of sphere, and r is the radius of ball.

Substitute 0.12m for r in the above equation.

$\begin{array}{c}v=\frac{4}{3}\pi {\left(0.12\text{\hspace{0.17em}m}\right)}^3\\ =7.238\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}}\end{array}$

The percentage of volume submerged is equal to 0.24 time of the volume of sphere. The force required to keep the 76% of ball inside the water will be,

$\begin{array}{l}F=\rho Vg-\rho \left(0.24V\right)g\\ F=0.76V\rho g\\ F=0.76\times \frac{4}{3}\pi r^3\times \rho \times g\end{array}$

Here, $\rho$is the density of water and g is the gravitational acceleration.

Substitute 0.12m for r , 1000kg/m³ for $\rho$ and 9.81m/s² for g in the above equation.

$\begin{array}{c}F=0.76\times \frac{4}{3}\pi {\left(0.12\text{\hspace{0.17em}m}\right)}^3\times 1000\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}\times 9.81\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\\ =53.96\text{\hspace{0.17em}N}\end{array}$

Hence, the required force is 53.96N .

Determination of the acceleration.

Part (b)

The relation of force in terms of mass and acceleration at the instant is expressed as,

$\begin{array}{c}F=ma\\ a=\frac{F}{m}\end{array}$

Here, m is the mass of ball is equal to the mass of water displaced when ball is floating, a is the acceleration.

The above equation can be written as

$\begin{array}{c}a=\frac{0.76v\rho g}{0.24v\rho }\\ a=\frac{0.76g}{0.24}\end{array}$

Substitute 9.81 m/s² for g in the above equation.

$\begin{array}{c}a=\frac{0.76\times 9.81\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}}{0.24}\\ =931.06\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\end{array}$

Hence, the value of acceleration is, 31.06 m/s² .