Chapter 8, Problem 75PQ

At 220 m, the bungee jump at the Verzasca Dam in Locarno, Switzerland, is one of the highest jumps on record. The length of the elastic cord, which can be modeled as having negligible mass and obeying Hooke’s law, has to be precisely tailored to each jumper because the margin of error at the bottom of the dam is less than 10.0 m. Kristin prepares for her jump by first hanging at rest from a 10.0-m length of the cord and is observed to stretch the rope to a total length of 12.5 m. a. What length of cord should Kristin use for her jump to be exactly 220 m? b. What is the maximum acceleration she will experience during her jump?

To determine

The length of the cord required to jump exactly $220\text{m}$

Answer to Problem 75PQ

The length of the cord required to jump exactly $220\text{m}$ is $\underset{\_}{110\text{m}}$.

Explanation of Solution

Given that the length of the test cord is $10\text{m}$ and it stretches $12.5\text{m}$. This indicates that the stretching length is $2.5\text{m}$.

Apply Hooke’s law to the rope. The stress on the rope is directly proportional to the strain.

  $F=kx$ (I)

Here, $F$ is the force or stress, $x$ is the strain, and $k$ is the force constant.

The weight of the person acts as the stress and the strain in the stretching length of the rope. Thus, equation (I) can be modified and solved for $k$ as,

  $\begin{array}{c}mg=kx\\ k=\frac{mg}{x}\end{array}$ (II)

Here, $m$ is the mass and $g$ is the acceleration due to gravity.

Since the stretch distance is $2.5\text{m}$ for the test cord, equation (II) can be modified as,

  $k=\frac{mg}{2.5\text{m}}$ (III)

The force constant for a longer cord is smaller and must depend inversely on the length of the cord For a length $L=10\text{m}$, the equation (III) can be expressed as,

  $\begin{array}{c}k=\frac{10\text{m}}{L}\frac{mg}{2.5\text{m}}\\ =4.00\frac{mg}{L}\end{array}$ (IV)

Equation (IV) can be used as a general expression for the force constant of the given cord of any length $L$.

Apply the law of conservation of energy at the top (initial condition) and bottom (final condition) of the drop.

  $K_i+U_{ig}+U_{is}=K_f+U_{fg}+U_{fs}$ (V)

Here, $K_i$ is the initial kinetic energy, $U_{ig}$ is the initial gravitational potential energy, $U_{is}$ is the initial elastic potential energy, $K_f$ is the final kinetic energy, $U_{fg}$ is the final gravitational potential energy, and $U_{fs}$ is the final elastic potential energy.

Write the general expression for the gravitational potential energy.

  $U_g=mgy$ (VI)

Here, $y$ is the height from ground.

Write the general expression for the elastic potential energy.

  $U_s=\frac{1}{2}k{x}^2$ (VII)

The initial and final kinetic energy of the person is zero. Moreover, the initial elastic potential energy is also zero. Thus, equation (V) can be modified using equations (VI) and (VII).

  $\begin{array}{c}0+mg{y}_i+0=0+mg{y}_f+\frac{1}{2}k{x}_f^2\\ mg\left(y_i-y_f\right)=\frac{1}{2}k{x}_f^2\end{array}$ (VII)

Here, $x_f$ is the stretched length of the cord during the dive, and $\left(y_i-y_f\right)$ represents the total distance covered in the jump.

Use equation (IV) in (VII).

  $mg\left(y_i-y_f\right)=\frac{1}{2}\left(4.00\frac{mg}{L}\right)x_f^2$ (VIII)

Since the total height of jump is $220\text{m}$, the stretched length of the cord during the dive can be evaluated as,

  $x_f=220\text{m}-L$ (IX)

Conclusion:

Use equation (IX) in (VIII) along with substituting $220\text{m}$ for $\left(y_i-y_f\right)$.

  $\begin{array}{c}mg\left(220\text{m}\right)=\frac{1}{2}\left(4.00\frac{mg}{L}\right){\left(220\text{m}-L\right)}^2\\ \left(220\text{m}\right)=\frac{1}{2}\left(\frac{4.00}{L}\right){\left(220\text{m}-L\right)}^2\\ \left(110\text{m}\right)L=L^2+\left(48400\text{m}\right)-\left(440\text{m}\right)L\\ 0=L^2-\left(550\text{m}\right)L+48400\text{m}\end{array}$ (X)

Solve the quadratic equation (X).

  $\begin{array}{c}L=\frac{550\text{m}\pm \sqrt{{\left(550\text{m}\right)}^2-4\left(1\right)\left(48400\text{m}\right)}}{2\left(1\right)}\\ =\frac{550\pm 330}{2}\\ =110\text{m or 880}\text{m}\end{array}$

Only the length which is less than $220\text{m}$ is physically meaningful, and hence the length $110\text{m}$ is the length of the cord.

Therefore, the length of the cord required to jump exactly $220\text{m}$ is $\underset{\_}{110\text{m}}$.

To determine

The maximum acceleration that the person experiences during the bungee jump.

Answer to Problem 75PQ

The maximum acceleration that the person experiences during the bungee jump is $\underset{\_}{29.4{\text{m/s}}^2}$.

Explanation of Solution

It is obtained that the length of the cord is $110\text{m}$.

During bungee jumping, the forces on the person are the gravitational force acting downward and the elastic restoring force acting upward. The total force acting on the person will be the difference of these two forces.

Write the expression for the total force acting on the person during the jump.

  ${\displaystyle \sum F}=+k{x}_{\max }-mg$ (XI)

Here, ${\displaystyle \sum F}$ is the total force, $x_{\max }$ is the maximum stretched length of the cord

Write the expression for the maximum stretched distance for the given cord.

  $x_{\max }=220\text{m}-L$ (XII)

Use Newton’s second law and equation (XII) to modify equation (XI).

  $ma=k\left(220\text{m}-L\right)-mg$ (XIII)

Here, $a$ is the acceleration of the person (since the maximum restoring force is considered, the acceleration will be the maximum).

Use equation (IV) in equation (XIII) and solve for $a$.

  $\begin{array}{c}ma=\left(4.00\frac{mg}{L}\right)\left(220\text{m}-L\right)-mg\\ a=g\left(\frac{4.00\left(220\text{m}-L\right)}{L}-1\right)\end{array}$ (XIV)

Conclusion:

Substitute $110\text{m}$ for $L$, and $9.8{\text{m/s}}^2$ for $g$ in equation (XIV) to find $a$.

  $\begin{array}{c}a=\left(9.8{\text{m/s}}^2\right)\left(\frac{4.00\left(220\text{m}-110\text{m}\right)}{110\text{m}}-1\right)\\ =29.4{\text{m/s}}^2\end{array}$

Therefore, the maximum acceleration that the person experiences during the bungee jump is $\underset{\_}{29.4{\text{m/s}}^2}$.