Chapter 3, Problem 3TC

Compare from greatest to least the accelerations of these skydivers.

(a) man with of air drag.

(b) woman with of air drag.

(c) woman with of air drag.

(d) dog with of air drag.

Greatest ___ ___ ___ ___Least

To determine

The comparison from greatest to least acceleration of skydivers.

Answer to Problem 3TC

The comparison from greatest to least acceleration of skydivers is $a_a>a_c>a_b>a_d$ .

Explanation of Solution

Given Info: Weight of the man in case (a) is 1000 N and air drag in case (a) is 800 N, weight of the woman in case (b) is 800 N and air drag in case (c) is 700 N, weight of the woman in case (c) is 700 N and air drag in case (c) is 600 N, weight of the dog in case (d) is 100 N and air drag in case (d) is 90 N.

Write the expression for net force on the man.

$F=W-F_{drag}$ …… (I)

Here,

$W$ is the weight of the man.

$F_{drag}$ is the air drag.

Substitute the value 1000N for W and 800N for $F_{drag}$ in the above expression to get the value of F.

$\begin{array}{l}{F}_a=1000\text{N}-800\text{N}\\ F_a=200\text{N}\end{array}$

Write the expression for the mass of the man.

$m=\frac{W}{g}$

Here,

g is the acceleration due to gravity.

Substitute the value 1000N for W and $10\text{m}/{\text{s}}^2$ for g in the above expression to get the value for m .

$\begin{array}{c}{m}_a=\frac{1000\text{N}}{9.8\text{m}/{\text{s}}^2}\\ m_a=\frac{1000\text{N}}{10\text{m}/{\text{s}}^2}\times \frac{1\text{kg}\cdot \text{m}/{\text{s}}^2}{1\text{N}}\\ m_a=100\text{kg}\end{array}$

Write the expression for the acceleration of the mass.

$a=\frac{F}{m}$

Substitute the value 200N for F and 100 kg for m in the above expression to get the value of a.

$\begin{array}{c}{a}_a=\frac{200\text{N}}{100\text{kg}}\\ a_a=\frac{200\text{N}}{100\text{kg}}\times \frac{1\text{kg}\cdot \text{m}/{\text{s}}^2}{1\text{N}}\\ a_a=2\text{m}/{\text{s}}^2\end{array}$

Thus, the acceleration of the man is $a_a=2\text{m}/{\text{s}}^2$.

Write the expression for the net force on the woman.

$F=W-F_{\text{drag}}$

Here,

$W$ is the weight of the woman.

$F_{\text{drag}}$ is the air drag.

Substitute the values 800N for W and 700N for F_drag in the above expression to get the value F.

$\begin{array}{l}{F}_b=800\text{N}-700\text{N}\\ F_b=100\text{N}\end{array}$

Write the expression for the mass of the woman.

$m=\frac{W}{g}$

Substitute $800\text{N}$ for $W$ and $10\text{m}/{\text{s}}^2$ for $g$ in the above expression to get $m$.

$\begin{array}{c}{m}_b=\frac{800\text{N}}{10\text{m}/{\text{s}}^2}\\ m_b=\frac{800\text{N}}{10\text{m}/{\text{s}}^2}\times \frac{1\text{kg}\cdot \text{m}/{\text{s}}^2}{1\text{N}}\\ m_b=80\text{kg}\end{array}$

Write the expression for the acceleration of the mass.

$a=\frac{F}{m}$

Substitute the values $200\text{N}$ for $F$ and $100\text{kg}$ for $m$ in the above expression to get the value $a$.

$\begin{array}{c}{a}_b=\frac{100\text{N}}{80\text{kg}}\\ a_b=\frac{100\text{N}}{80\text{kg}}\times \frac{1\text{kg}\cdot \text{m}/{\text{s}}^2}{1\text{N}}\\ a_b=1.25\text{m}/{\text{s}}^2\end{array}$

Thus, the acceleration of the man is $a_b=1.25\text{m}/{\text{s}}^2$.

Write the expression for the net force on the man.

$F=W-F_{\text{drag}}$

Here,

$W$ is the weight of the man.

$F_{drag}$ is the air drag.

Substitute the values $700\text{N}$ for $W$ and $600\text{N}$ for $F_{\text{drag}}$ in the above expression to get the value of $F$.

$\begin{array}{l}{F}_c=700\text{N}-600\text{N}\\ F_c=100\text{N}\end{array}$

Write the expression for the mass of the woman.

$m=\frac{W}{g}$

Substitute 700N for W and 10m/s² for g in the above expression to get m.

$\begin{array}{c}{m}_c=\frac{700\text{N}}{9.8\text{m}/{\text{s}}^2}\\ m_c=\frac{700\text{N}}{10\text{m}/{\text{s}}^2}\times \frac{1\text{kg}\cdot \text{m}/{\text{s}}^2}{1\text{N}}\\ m_c=70\text{kg}\end{array}$

Write the expression for the acceleration of the mass.

$a=\frac{F}{m}$

Substitute $200\text{N}$ for $F$ and $100\text{kg}$ for $m$ in the above expression to get $a$.

$\begin{array}{c}{a}_c=\frac{100\text{N}}{100\text{kg}}\\ a_c=\frac{100\text{N}}{70\text{kg}}\times \frac{1\text{kgm}/{\text{s}}^2}{1\text{N}}\\ a_c=1.4\text{m}/{\text{s}}^2\end{array}$

Thus, the acceleration of the man is $a_c=1.4\text{m}/{\text{s}}^2$.

Write the expression for net force on the man.

$F=W-F_{drag}$

Here,

$W$ is the weight of the man.

$F_{\text{drag}}$ is the air drag.

Substitute the values$100\text{N}$ for $W$ and $90\text{N}$ for $F_{\text{drag}}$ in the above expression to get the value of $F$ .

$\begin{array}{l}{F}_d=700\text{N}-600\text{N}\\ F_d=100\text{N}\text{.}\end{array}$

Write the expression for the mass of the dog.

$m=\frac{W}{g}.$

Substitute the values $700\text{N}$ for $W$ and $10\text{m}/{\text{s}}^2$ for $g$ in the above expression to get the value $m$ .

$\begin{array}{c}{m}_d=\frac{100\text{N}}{9.8\text{m}/{\text{s}}^2}\\ m_d=\frac{100\text{N}}{10\text{m}/{\text{s}}^2}\times \frac{1\text{kgm}/{\text{s}}^2}{1\text{N}}\\ m_d=10\text{kg}\end{array}$

Write the expression for the acceleration of the mass.

$a=\frac{F}{m}$

Substitute the values $10\text{N}$ for $F$ and $10\text{kg}$ for $m$ in the above expression to get the value of $a$ .

$\begin{array}{c}{a}_d=\frac{10\text{N}}{10\text{kg}}\\ a_d=\frac{10\text{N}}{10\text{kg}}\times \frac{1\text{kg}\cdot \text{m}/{\text{s}}^2}{1\text{N}}\\ a_d=1\text{m}/{\text{s}}^2\end{array}$

Thus, the acceleration of the man is $a_d=1\text{m}/{\text{s}}^2$ .

Conclusion:

Therefore, the ranking from greatest to least of the acceleration of skydivers is .