Chapter 6, Problem 110P

(a)

To determine

Total work done on the stunt woman during the fall.

Answer to Problem 110P

The work done on the stunt woman during the fall is $\underset{\_}{3.45\text{kJ}}$.

Explanation of Solution

Write the work energy theorem

    $W_{\text{total}}=\Delta K$        (I)

Here $W_{\text{total}}$ is the total work done and $\Delta K$ is the change in kinetic energy.

Write the expression for kinetic energy.

    $\Delta K=K_f-K_i$        (II)

Here $K_f$ is the final kinetic energy $K_i$ is the initial kinetic energy.

The initial kinetic energy of the woman is zero.

Then,

    $\begin{array}{c}{W}_{\text{total}}=K_f\\ =\frac{1}{2}m{v}_f{}^2\end{array}$        (III)

Here $m$ is the mass of the woman and $v_f$ is the final velocity.

Conclusion

Substitute $62.5\text{kg}$ for $m$ and $\text{10}\text{.5}\text{m}/\text{s}$ for $v_f$ in (III)

    $\begin{array}{c}{W}_{\text{total}}=\frac{1}{2}\left(62.5\text{kg}\right){\left(\text{10}\text{.5}\text{m}/\text{s}\right)}^2\\ =3.45\text{kJ}\end{array}$

The work done on the stunt woman during the fall is $\underset{\_}{3.45\text{kJ}}$.

(b)

To determine

The work done by gravity on the stunt woman.

Answer to Problem 110P

The work done by gravity on the stunt woman is $\underset{\_}{4.96\text{kJ}}$.

Explanation of Solution

The work done by gravity is the negative of the change in gravitational potential energy of the stunt woman.

Write the expression for work done

    $W_{\text{grav}}=\Delta U$        (IV)

Here $W_{\text{grav}}$ is the work done by the gravity on the stunt woman and $\Delta U$ is the change in potential energy of the stunt woman.

Write the equation for $\Delta U$

    $\Delta U=mg\Delta y$        (V)

Here $g$ is the acceleration due to gravity and $\Delta y$ is the height through which the stunt woman has fallen.

Substitute (IV) in (V)

    $W_{\text{grav}}=mg\Delta y$        (VI)

Conclusion

Substitute $62.5\text{kg}$ for $m$, $\text{9}\text{.8}\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $-8.10\text{m}$ for $\Delta y$ in (VI)

    $\begin{array}{c}{W}_{\text{grav}}=\left(62.5\text{kg}\right)\left(\text{9}\text{.8}\text{m}/{\text{s}}^{\text{2}}\right)\left(8.10\text{m}\right)\\ =4.96\text{kJ}\end{array}$

The work done by gravity on the stunt woman is $\underset{\_}{4.96\text{kJ}}$.

(c)

To determine

The work done by air resistance on the stunt woman.

Answer to Problem 110P

The work done by air resistance on the stunt woman is $\underset{\_}{-1.52\text{kJ}}$.

Explanation of Solution

Write the equation for total work done.

    $W_{\text{total}}=W_{\text{grav}}+W_{\text{air}}$        (VII)

Here $W_{\text{air}}$ is the work done by air resistance on the stunt woman.

Rewrite (VII) in terms of $W_{\text{air}}$.

    $W_{\text{air}}=W_{\text{total}}-W_{\text{grav}}$        (VIII)

Conclusion0.

Substitute $3.45\text{kJ}$ for $W_{\text{total}}$ and $4.96\text{kJ}$ for $W_{\text{grav}}$ in (VIII)

    $\begin{array}{c}{W}_{\text{air}}=3.45\text{kJ}-4.96\text{kJ}\\ =-1.52\text{kJ}\end{array}$

The work done by air resistance on the stunt woman is $\underset{\_}{-1.52\text{kJ}}$.

(d)

To determine

Magnitude of average force of air resistance.

Answer to Problem 110P

The magnitude of average force of air resistance is $\underset{\_}{187\text{N}}$.

Explanation of Solution

Take the positive y direction to be upward.

Write the equation for work done by air resistance.

    $W_{\text{air}}=F_{\text{air}}\Delta y$        (IX)

Here $F_{\text{air}}$ is the force of air resistance

Rewrite (IX) in terms of $F_{\text{air}}$

    $F_{\text{air}}=\frac{W_{\text{air}}}{\Delta y}$        (X)

Conclusion

Substitute $-1.52\text{kJ}$ for $W_{\text{air}}$ and $-8.10\text{m}$ for $\Delta y$ in (X)

    $\begin{array}{c}{F}_{\text{air}}=\frac{-1.52\text{kJ}}{-8.10\text{m}}\\ \approx 187\text{N}\end{array}$

The magnitude of average force of air resistance is $\underset{\_}{187\text{N}}$.