Chapter 6, Problem 28P

(a)

To determine

The skydiver acceleration when her speed is $30.0\text{m}/\text{s}$.

Answer to Problem 28P

The skydiver acceleration when her speed is $30.0\text{m}/\text{s}$ is $\underset{\_}{6.27\text{m}/{\text{s}}^{\text{2}}}$ in a downward direction.

Explanation of Solution

When an object passes through the air it experiences drag force due to the air. The drag force is directly proportional to the square of the velocity of the object.

Write the expression for resistive force by the air on the skydiver as.

  $R=D{v}^2$                                                                                                         (I)

Here, $R$ is a restive force, $D$ is drag coefficient and $v$ is the speed of skydiver.

The forces act on the skydiver as shown below.

Write the expression corresponding to Newton’s second law of motion in $y$-direction as.

  $mg-R=ma$                                                                                                (II)

Here, $m$ is mass of the skydiver, $a$ is an acceleration of the skydiver and $g$ is gravitational acceleration.

Substitute $D{v}^2$ for $R$ in equation (II).

  $mg-D{v}^2=ma$

Simplify the equation (II) for $a$ as.

  $a=\frac{mg-D{v}^2}{m}$                                                                                            (III)

At the terminal speed of the skydiver, the acceleration becomes zero. If the acceleration is zero then the velocity becomes terminal velocity.

Substitute $0\text{m}/{\text{s}}^{\text{2}}$ for $a$ and $D{v}_T{}^2$ for $R$ in equation (II).

  $\begin{array}{c}mg-\left(D{v}_T{}^2\right)=m\left(0\text{m}/{\text{s}}^{\text{2}}\right)\\ mg-D{v}_T{}^2=0\\ D{v}_T{}^2=mg\end{array}$

Simplify the above expression for $D$ as.

  $D=\frac{mg}{v_T{}^2}$                                                                                                   (IV)

Here, $v_T$ is terminal velocity of the skydiver.

Conclusion:

Substitute $80.0\text{kg}$ for $m$, $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $50\text{m}/\text{s}$ for $v_T$ in equation (IV).

  $\begin{array}{c}D=\frac{\left(80.0\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)}{{\left(50\text{m}/\text{s}\right)}^2}\\ =0.314\text{N}\cdot {\text{s}}^{\text{2}}/{\text{m}}^{\text{2}}\end{array}$

Substitute $80.0\text{kg}$ for $m$, $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$, $0.314\text{N}\cdot {\text{s}}^{\text{2}}/{\text{m}}^{\text{2}}$ for $D$ and $30\text{m}/\text{s}$ for $v$ in equation (III).

  $\begin{array}{c}a=\frac{\left(80.0\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)-\left(0.314\text{N}\cdot {\text{s}}^{\text{2}}/{\text{m}}^{\text{2}}\right){\left(30\text{m}/\text{s}\right)}^2}{\left(80.0\text{kg}\right)}\\ =\frac{784\text{N}-9.42\text{N}}{\left(80.0\text{kg}\right)}\\ =6.267\text{m}/{\text{s}}^{\text{2}}\\ \approx 6.27\text{m}/{\text{s}}^{\text{2}}\end{array}$

The positive direction of the acceleration indicates downward direction.

Thus, the skydiver acceleration when her speed is $30.0\text{m}/\text{s}$ is $\underset{\_}{6.27\text{m}/{\text{s}}^{\text{2}}}$ in a downward direction.

(b)

To determine

The drag force on the skydiver when her speed is $50\text{m}/\text{s}$.

Answer to Problem 28P

The drag force on the skydiver when her speed is $50\text{m}/\text{s}$ is $784\text{N}$ acts in an upward direction.

Explanation of Solution

At the terminal speed, the weight of the skydiver becomes equal to the drag force.

Write the expression corresponding to the condition of equilibrium in $y$-direction as.

  $R=mg$                                                                                                         (V)

Conclusion:

Substitute $80.0\text{kg}$ for $m$ and $9.80\text{m}/{\text{s}}^{\text{2}}$ for  $g$ in equation (V).

  $\begin{array}{c}R=\left(80.0\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\\ =784\text{N}\end{array}$

The positive sign indicates an upward direction of the force.

Thus, the drag force on the skydiver when her speed is $50\text{m}/\text{s}$ is $784\text{N}$ acts in an upward direction.

(c)

To determine

The drag force on the skydiver when her speed is $30\text{m}/\text{s}$.

Answer to Problem 28P

The drag force on the skydiver at a speed of $30\text{m}/\text{s}$ is $283\text{N}$ and it acts in an upward direction.

Explanation of Solution

Conclusion:

Substitute $0.314\text{N}\cdot {\text{s}}^{\text{2}}/{\text{m}}^{\text{2}}$ for $D$ and $30\text{m}/\text{s}$ for $v$ in equation (I).

  $\begin{array}{c}R=\left(0.314\text{N}\cdot {\text{s}}^{\text{2}}/{\text{m}}^{\text{2}}\right){\left(30\text{m}/\text{s}\right)}^2\\ =283\text{N}\end{array}$

The positive sign of the resistive force indicates an upward direction of the force.

Thus, the drag force on the skydiver at a speed of $30\text{m}/\text{s}$ is $283\text{N}$ and it acts in an upward direction.