Chapter 6, Problem 111P

(a)

To determine

The total work done on the stunt woman during the fall.

Answer to Problem 111P

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

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 $\Delta K$ .

$\Delta K=K_f-K_i$

Here, $K_f$ is the final kinetic energy and $K_i$ is the initial kinetic energy

The initial kinetic energy of the woman is zero.

Substitute $0$ for $K_i$ in the above equation.

$\begin{array}{c}\Delta K=K_f-0\\ \Delta K=K_f\end{array}$ (II)

Write the expression for $K_f$ .

$K_f=\frac{1}{2}m{v}_f^2$

Here, $m$ is the mass of the stunt woman and $v_f$ is her final speed

Put the above equation in equation (II).

$\Delta K=\frac{1}{2}m{v}_f^2$

Put the above equation in equation (I).

$W_{\text{total}}=\frac{1}{2}m{v}_f^2$ (III)

Conclusion:

Given that the mass of the stunt woman is $62.5\text{ kg}$ and her final speed is $10.5\text{ m/s}$ .

Substitute $62.5\text{ kg}$ for $m$ and $10.5\text{ m/s}$ for $v_f$ in equation (III) to find the value of $W_{\text{total}}$.

$\begin{array}{c}{W}_{\text{total}}=\frac{1}{2}\left(62.5\text{ kg}\right){\left(10.5\text{ m/s}\right)}^2\\ =3.45\times {10}^3\text{ J}\\ =3.45\text{ kJ}\end{array}$

Therefore, the total work done on the stunt woman during the fall is $3.45\text{ kJ}$ .

(b)

To determine

Thework done by gravity on the stunt woman.

Answer to Problem 111P

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

Explanation of Solution

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

$W_{\text{grav}}=-\Delta U$ (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 $\Delta U$ .

$\Delta U=mg\Delta y$

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

Put the above equation in equation (IV).

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

Conclusion:

Given that the height through which the stunt woman has fallen is $8.10\text{ m}$ . The value of acceleration due to gravity is $9.80{\text{ m/s}}^2$ .

Substitute $62.5\text{ kg}$ for $m$, $9.80{\text{ m/s}}^2$ for $g$ and $-8.10\text{ m}$ for $\Delta y$ in equation (V) to find $W_{\text{grav}}$.

$\begin{array}{c}{W}_{\text{grav}}=-\left(62.5\text{ kg}\right)\left(9.80{\text{ m/s}}^2\right)\left(-8.10\text{ m}\right)\\ =4.96\times {10}^3\text{ J}\\ =4.96\text{ kJ}\end{array}$

Therefore, the work done by gravity on the stunt woman is $4.96\text{ kJ}$ .

(c)

To determine

Thework done by air resistance on the stunt woman.

Answer to Problem 111P

Thework done by air resistance on the stunt woman is $-1.52\text{ kJ}$ .

Explanation of Solution

The work is done on the stunt woman by gravity as well as air resistance.

$W_{\text{total}}=W_{\text{grav}}+W_{\text{air}}$

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

Rewrite the above equation for $W_{\text{air}}$ .

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

Conclusion:

Substitute $3.45\text{ kJ}$ for $W_{\text{total}}$ and $4.96\text{ kJ}$ for $W_{\text{grav}}$ in equation (VI) to find $W_{\text{air}}$ .

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

Therefore, the work done by air resistance on the stunt woman is $-1.52\text{ kJ}$ .

(d)

To determine

The magnitude of the average force of air resistance.

Answer to Problem 111P

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

Explanation of Solution

Take the $+y$ direction to be upward.

Write the above equation for $W_{\text{air}}$ .

$W_{\text{air}}=F_{\text{air}}\Delta y$

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

Rewrite the above equation for $F_{\text{air}}$ .

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

Conclusion:

Substitute $-1.52\text{ kJ}$ for $W_{\text{air}}$ and $-8.10\text{ m}$ for $\Delta y$ in equation (VII) to find $F_{\text{air}}$ .

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

Therefore, the magnitude of the average force of air resistance is $187\text{ N}$ .