Chapter 6, Problem 116P

(a)

To determine

The distance moved by person J to reach the lowest point.

Answer to Problem 116P

The distance moved by person J to reach the lowest point is $h=L-L\cos 20°$.

Explanation of Solution

Figure 1 represents the motion of person J.

Here, $L$ is the length of the vine and $h$ is the vertical displacement of person J

Conclusion:

Refer Figure 1.

Write an expression for $L$.

$L=h+L\cos 20°$

Rewrite equation (I) to find $h$.

$h=L-L\cos 20°$

Thus, the distance moved by person J to reach the lowest point is $h=L-L\cos 20°$.

(b)

To determine

The speed of person J at the lowest point.

Answer to Problem 116P

The speed of person J at the lowest point is $4.9\text{m/s}$.

Explanation of Solution

Write an expression to calculate the conservation of mechanical energy of person J.

               $\begin{array}{c}{K}_i+U_i=K_f+U_f\\ \frac{1}{2}m{v}_i^2+mg{h}_i=\frac{1}{2}m{v}_f^2+mg{h}_f\\ \frac{1}{2}{v}_i^2+g{h}_i=\frac{1}{2}{v}_f^2+g{h}_f\\ \frac{1}{2}{v}_i^2+g\left(h\right)=\frac{1}{2}{v}_f^2+g\left(0\right)\end{array}$

$\begin{array}{c}\frac{1}{2}{v}_i^2+gh=\frac{1}{2}{v}_f^2\\ \frac{1}{2}{v}_i^2+g\left(L-L\cos 20°\right)=\frac{1}{2}{v}_f^2\\ \frac{1}{2}{v}_i^2+gL\left(1-\cos 20°\right)=\frac{1}{2}{v}_f^2\end{array}$ (I)

Here, $K_i$ is the initial kinetic energy of the person J, $U_i$ is the initial potential energy, $K_f$ is the final kinetic energy of the person, $U_f$ is the final potential energy of the person, $m$ is the mass of the person, $g$ is the acceleration due to gravity, $h_i$ is the initial height of the person, $h_f$ is the final height of the person, $v_i$ is the initial velocity of the person and $v_f$ is the final velocity of the person.

Rearrange the equation (I) to find $v_f$.

$v_f=\sqrt{v_i^2+2gL\left(1-\cos 20°\right)}$ (II)

Conclusion:

Substitute $7.0\text{m}$ for $L$, $4.0\text{m/s}$ for $v_i$ and $9.8{\text{m/s}}^{\text{2}}$ for $g$ in equation (II) to find $v_f$.

$\begin{array}{c}{v}_f=\sqrt{{\left(4.0\text{m/s}\right)}^2+2\left(9.8{\text{m/s}}^{\text{2}}\right)\left(7.0\text{m}\right)\left(1-\cos 20°\right)}\\ =\sqrt{\left(16{\text{m}}^2{\text{/s}}^2\right)+\left(8.2{\text{m}}^2{\text{/s}}^2\right)}\\ =4.9\text{m/s}\end{array}$

Thus, the speed of person J at the lowest point is $4.9\text{m/s}$.

(c)

To determine

The height at which person J can swing about the lowest point.

Answer to Problem 116P

The height at which person J can swing about the lowest point is $1.24\text{m}$.

Explanation of Solution

Write an expression to calculate the conservation of mechanical energy of person J.

$\begin{array}{c}{K}_i+U_i=K_f+U_f\\ \frac{1}{2}m{v}_i^2+mg{h}_i=\frac{1}{2}m{v}_f^2+mg{h}_f\\ \frac{1}{2}{v}_i^2+g{h}_i=\frac{1}{2}{v}_f^2+g{h}_f\\ \frac{1}{2}{v}_i^2+g\left(h\right)=\frac{1}{2}{\left(0\right)}^2+g{h}_f\\ \frac{1}{2}{v}_i^2+gh=g{h}_f\\ \frac{1}{2}{v}_i^2+g\left(L-L\cos 20°\right)=g{h}_f\\ \frac{1}{2}{v}_i^2+gL\left(1-\cos 20°\right)=g{h}_f\end{array}$ (III)

Rearrange the equation (III) to find $h_f$.

$h_f=\frac{1}{2g}{v}_i^2+L\left(1-\cos 20°\right)$ (IV)

Conclusion:

Substitute $7.0\text{m}$ for $L$, $4.0\text{m/s}$ for $v_i$ and $9.8{\text{m/s}}^{\text{2}}$ for $g$ in equation (IV) to find $h_f$.

$\begin{array}{c}{h}_f=\frac{1}{2\left(9.8{\text{m/s}}^{\text{2}}\right)}{\left(4.0\text{m/s}\right)}^2+\left(7.0\text{m}\right)\left(1-\cos 20°\right)\\ =0.81\text{m}+0.42\text{m}\\ =\text{1}\text{.24}\text{m}\end{array}$

Thus, the height at which person J can swing about the lowest point is $1.24\text{m}$.