Chapter 8, Problem 8.81CP

To determine

The minimum speed of the swing with which Jane must begin her swing to just make it to the other side.

Answer to Problem 8.81CP

The minimum speed of the swing with which Jane must begin her swing to just make it to the other side is $6.15\text{m}/\text{s}$.

Explanation of Solution

Given info: The mass of Jane is $50.0\text{kg}$, magnitude of force is $110\text{}\text{N}$, mass of Tarzan is $80.0\text{kg}$, length of vine  is $40.0\text{m}$, distance between Tarzan and Jane is $D$.

The diagram is shown below.

Figure I

From the figure, the width of the river is,

$D=L\sin \varphi +L\sin \theta$

Here,

$L$ is the length of the vine.

$\theta$ is the angle between vine and the vertical at initial state.

$\varphi$ is the angle between the vine and vertical at the final state.

Rearrange the above formula for $\varphi$.

$\begin{array}{c}D=L\sin \varphi +L\sin \theta \\ D-L\sin \theta =L\sin \varphi \\ \sin \varphi =\frac{D-L\sin \theta }{L}\\ \varphi ={\sin }^{-1}\left(\frac{D-L\sin \theta }{L}\right)\end{array}$

Substitute $50.0\text{m}$ for $D$, $40.0\text{m}$ for $L$, $50°$ for $\theta$ in the above formula to find $\varphi$.

$\begin{array}{c}\varphi ={\sin }^{-1}\left(\frac{D-L\sin \theta }{L}\right)\\ ={\sin }^{-1}\left(\frac{50.0\text{m}-\left(40.0\text{m}\right)\text{sin50}°}{50.0\text{m}}\right)\\ ={\sin }^{-1}\left(0.484\right)\\ =28.94°\end{array}$

Thus, the value of $\varphi$ is $28.94°$.

The formula to calculate the initial kinetic energy of Jane is,

$K.E_1=\frac{1}{2}m{v}_1{}^2$

Here,

$m$ is the mass of the Jane.

$v_1$ is the initial velocity of Jane.

Thus, the initial kinetic energy of Jane is $\frac{1}{2}m{v}_1{}^2$.

The formula to calculate the final kinetic energy of Jane is,

$K.E_2=\frac{1}{2}m{v}_2{}^2$

Here,

$m$ is the mass of the person.

$v_2$ is the final velocity of the person.

Substitute $0$ for $v_2$ in the above formula to find $K.E_2$.

$\begin{array}{c}K.E_2=\frac{1}{2}m{v}_2{}^2\\ =\frac{1}{2}m{\left(0\right)}^2\\ =0\text{J}\end{array}$

Thus, the final kinetic energy of the Jane is $0\text{}\text{J}$.

The formula to calculate the initial gravitational potential energy is,

$P.E_1=mg\left(-L\cos \theta \right)$

Here,

$m$ is the mass of the Jane.

$\theta$ is the angle between the vine and vertical at the initial state.

$g$ is the acceleration due to gravity.

Thus, the initial gravitational potential energy is $-mgL\cos \theta$.

The formula to calculate the final gravitational potential energy is,

$P.E_2=mg\left(-L\cos \varphi \right)$

Here,

$m$ is the mass of the Jane.

$\varphi$ is the angle between the vine and vertical at final state.

$g$ is the acceleration due to gravity.

Thus the final potential energy of the car is $-mgL\cos \varphi$.

The formula to calculate the initial work done of the wind due to constant force is,

$W=FD\cos 180°$

Here,

$F$ is the constant horizontal force.

$D$ is the width of the river.

Substitute $-1$ for $\cos 180°$ in the above formula to find $W$.

$\begin{array}{c}W=FD\cos 180°\\ =FD\left(-1\right)\\ =-FD\end{array}$

Thus, the initial work done of the wind is $-FD$.

The formula to calculate the law of conservation of energy to the total system is,

$K.E_1+U_1+W=K.E_2+U_2$

Here,

$K.E_1$ is the initial kinetic energy of Jane.

$K.E_2$ is the final kinetic energy of Jane.

$U{1}_{}$ is the initial gravitational potential energy.

$U_2$ is the final gravitational potential energy.

$W$ is the initial work done of the wind.

Substitute $\frac{1}{2}m{v}_1{}^2$ for $K.E_1$ and $0$ for $K.E_2$ , $-FD$ for $W$ , $-mgL\cos \theta$ for $U_1$ , $-mgL\cos \varphi$ for $U_2$  in the above formula to find $v_1$.

$\begin{array}{c}K.E_1+U_1+W=K.E_2+U_2\\ \frac{1}{2}m{v}_1{}^2+-mgL\cos \theta +\left(-FD\right)=0+\left(-mgL\cos \varphi \right)\\ v_1=\sqrt{2gL\cos \theta +\frac{2FD}{m}-mgL\cos \varphi }\end{array}$

Substitute $110\text{N}$ for $F$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $50°$ for $\theta$, $28.94°$ for $\varphi$, $50.0\text{kg}$ for $m$ in the above formula to find $v_1$.

$\begin{array}{c}{v}_1=\sqrt{2gL\cos \theta +\frac{2FD}{m}-mgL\cos \varphi }\\ =\sqrt{2\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(40.0\text{m}\right)\cos 50°+\frac{2\left(110\text{N}\right)\left(50.0\text{m}\right)}{\left(50.0\text{kg}\right)}-\left(50.0\text{}\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(40.0\text{}\text{m}\right)\cos 28.94°}\\ =6.15\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the minimum speed of the swing that must Jane begin her swing to just make it to the other side is $6.15\text{m}/\text{s}$.

To determine

The minimum speed of the swing at the beginning.

Answer to Problem 8.81CP

The minimum speed of the swing at the beginning is $9.87\text{}\text{m}/\text{s}$

Explanation of Solution

Given info: The mass of Jane is $50.0\text{kg}$ , magnitude of force is $110\text{}\text{N}$, mass of Tarzan  is $80.0\text{kg}$, length of vine  is $40.0\text{m}$, distance between Tarzan and Jane is $D$.

The formula to calculate the combined mass is,

$M_{\text{c}}=m_{\text{j}}+m_{\text{t}}$

Here,

$m_{\text{j}}$ is the mass of Jane.

$m_{\text{t}}$ is the mass of Tarzan.

Substitute $50.0\text{kg}$ for $m_{\text{j}}$ and $80.0\text{kg}$ for $m_{\text{t}}$ in the above formula to find $M_{\text{c}}$.

$\begin{array}{c}{M}_{\text{c}}=m_{\text{j}}+m_{\text{t}}\\ =\left(50.0\text{kg}\right)+\left(80.0\text{kg}\right)\\ =130.0\text{kg}\end{array}$

The formula to calculate the initial kinetic energy is,

$K.E_1=\frac{1}{2}{M}_{\text{c}}{v}_1{}^2$

Here,

$M_{\text{c}}$ is the combined mass.

$v_1$ is the initial velocity.

Thus, the initial kinetic energy is $\frac{1}{2}{M}_{\text{c}}{v}_1{}^2$.

The formula to calculate the final kinetic energy of Jane is,

$K.E_2=\frac{1}{2}{M}_{\text{c}}{v}_2{}^2$

Here,

$M_{\text{c}}$ is the combined mass.

$v_2$ is the final velocity .

Substitute $0$ for $v_2$ in the above formula to find $K.E_2$.

$\begin{array}{c}K.E_2=\frac{1}{2}{M}_{\text{c}}{v}_2{}^2\\ =\frac{1}{2}{M}_{\text{c}}{\left(0\right)}^2\\ =0\text{J}\end{array}$

Thus, the final kinetic energy is $0\text{}\text{J}$.

The formula to calculate the initial gravitational potential energy is,

$P.E_1=M_{\text{c}}g\left(-L\cos \varphi \right)$

Here,

$M_{\text{c}}$ is the combined mass.

$\varphi$ is the angle between the vine and vertical at the final state.

$g$ is the acceleration due to gravity.

Thus, the initial gravitational potential energy is $-M_{\text{c}}gL\cos \varphi$.

The formula to calculate the final gravitational potential energy is,

$P.E_2=M_{\text{c}}g\left(-L\cos \theta \right)$

Here,

$M_{\text{c}}$ is the combined mass.

$\theta$ is the angle between the vine and vertical at initial state.

$g$ is the acceleration due to gravity.

Thus the final potential energy of the car is $-M_{\text{c}}gL\cos \theta$.

The formula to calculate the initial work done of the wind due to constant force is,

$W=FD\cos 0°$

Here,

$F$ is the constant horizontal force.

$D$ is the width of the river.

Substitute $1$ for $\cos 0°$ in the above formula to find $W$.

$\begin{array}{c}W=FD\cos 0°\\ =FD\left(1\right)\\ =FD\end{array}$

Thus, the initial work done of the wind is $FD$.

The formula to calculate the law of conservation of energy to the total system is,

$K.E_1+U_1+W=K.E_2+U_2$

Here,

$K.E_1$ is the initial kinetic energy of Jane.

$K.E_2$ is the final kinetic energy of Jane.

$U{1}_{}$ is the initial gravitational potential energy.

$U_2$ is the final gravitational potential energy.

$W$ is the initial work done of the wind.

Substitute $\frac{1}{2}{M}_{\text{c}}{v}_1{}^2$ for $K.E_1$ and $0$ for $K.E_2$ , $FD$ for $W$ , $-M_{\text{c}}gL\cos \varphi$ for $U_1$ , $-M_{\text{c}}gL\cos \theta$ for $U_2$ in the above formula to find $v_1$.

$\begin{array}{c}K.E_1+U_1+W=K.E_2+U_2\\ \frac{1}{2}{M}_{\text{c}}{v}_1{}^2+-M_{\text{c}}gL\cos \varphi +\left(FD\right)=0+\left(-M_{\text{c}}gL\cos \theta \right)\\ v_1=\sqrt{2gL\cos \varphi -\frac{FD}{M_{\text{c}}}-2gL\cos \theta }\end{array}$

Substitute $110\text{N}$ for $F$ , $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $50°$ for $\theta$, $28.94°$ for $\varphi$, $130.0\text{kg}$ for $M_{\text{c}}$ in the above formula to find $v_1$.

$\begin{array}{c}{v}_1=\sqrt{2gL\cos \varphi -\frac{FD}{M_{\text{c}}}-2gL\cos \theta }\\ =\sqrt{2\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(40.0\text{m}\right)\cos 28.94°-\frac{\left(110\text{N}\right)\left(50.0\text{m}\right)}{\left(130.0\text{kg}\right)}-\left(50.0\text{}\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(40.0\text{}\text{m}\right)\cos 50°}\\ =9.87\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the minimum speed of the swing at the beginning is $9.87\text{}\text{m}/\text{s}$.