Question: Design a “bungee jump” apparatus for adults. A bungee jumper falls from a high platform with two elastic cords tied to the ankles. The jumper falls freely for a while, with the cords slack. Then the jumper falls an additional distance with the cords increasingly tense. You have cords that are10m long, and these cords stretch in the jump an additional 24mfor a jumper whose mass is 80kg, the heaviest adult you will allow to use your bungee jump (heavier customers would hit the ground). You can neglect air resistance. (a) Make a series of five simple diagrams, like a comic strip, showing the platform, the jumper, and the two cords at various times in the fall and the rebound. On each diagram, draw and label vectors representing the forces acting on the jumper, and the jumper’s velocity. Make the relative lengths of the vectors reflect their relative magnitudes. (b) At what instant is there the greatest tension in the cords? How do you know? (c) What is the jumper’s speed at this instant? (d) Is the jumper’s momentum changing at this instant or not? (That is, is$\mathrm{d}\stackrel{\rightharpoonup }{\mathrm{p}}/dt$nonzero or zero?) Explain briefly. (e) Focus on this instant, and use the principles of this chapter to determine the spring stiffness$k_s$for each cord. Explain your analysis. (f) What is the maximum tension that each cord must support without breaking? (g) What is the maximum acceleration (in g’s) that the jumper experiences? What is the direction of this maximum acceleration? (h) State clearly what approximations and estimates you have made in your design.

The given data is listed below as,

• The length of the cords are h = 10m.
• The cords can be stretched to an additional length of x = 24m.
• The mass of the jumper is m = 80kg.

Hooke’s law states that the deforming force of a body is directly proportional to the displacement of the body and the spring constant.

The law of energy conservation states that the energy of an isolated system always remains constant irrespective of the internal changes.

Newton’s second law states that the force exerted on an object is directly proportional to the product of the mass and the acceleration of an object.

The five simple diagrams have been provided below along with the jumper’s velocity and the label vector.

Here, in the diagram A, mg is the downward force and $v_A$is the velocity of the body. In the diagram B,$v_B$is the velocity of the body and $k{s}_B$is the upward force. In the diagram C and D,$k{s}_C$and $k{s}_D$is the upward force and$v_C$ and$v_D$ are the velocity of the body. In the diagram E,$v_E$ is the velocity.

The tension is greatest when the bungee jumper is at the furthest point from the equilibrium position. However, it has been identified that the cord has been stretched to an additional 24m. Hence, according to the Hooke’s law, the tension is greatest at this point as the tension is directly proportional to the displacement of the cords.

Thus, as the jumper is at the furthest point from the equilibrium position, that produces the greatest tension in the chords.

For making the greatest tension, the jumper needs to stay at the minimum possible altitude that is the height should be zero by setting the reference of the gravitational potential energy at the lowest possible point. Moreover, for making this happen, the speed also needs to be zero.

The equation of the speed and the height of the jumper is expressed as:

$\frac{1}{2}k{s}_f^2=mg(h_i-h_f)-\frac{1}{2}m{v}_f^2$

Here, k is the spring constant,$s_f$ is the displacement, m is the mass, g is the gravitational acceleration, h is the initial and$h_f$ is the final height and v is the final velocity.

If the final height and the velocity is zero, then only the energy is conserved.

${\left(\frac{1}{2}k{s}_f^2\right)}_{\max }=mg{h}_i$

Thus, the jumper’s speed at this instant is 0 .

The value of$d\overline{p}ldt$ can be written as$mdVldt$ as the momentum is the product of the mass and the velocity of the jumper The jumper’s momentum is changing due to the reason that the velocity of the jumper is continuously changing. So, the value of$d\overline{p}ldt$ is not 0 .

Thus, the jumper’s momentum is changing at this instant.

The equation of the potential energy due to gravity can be expressed as:

$P{E}_g=mgh$

Here, $P{E}_g$is the potential energy due to gravity which is the product of the mass of the jumper m and acceleration due to gravity g and h the cord length.

The equation of the upward potential energy is:

$PE=-\frac{1}{2}{k}_s{x}^2$

The upward potential energy of the jumper$PE$that is the half of the product of the stiffness of the spring$k_s$ and square of the displacement x.

The equation of the potential energy at the bottom is:

$P{E}_g^f=mgx$

The potential energy at the bottom $P{E}_g^f$is the product of the mass of the jumper m, acceleration due to gravity g and the downward stretching of the cord x.

The equation of the downward potential energy is:

$P{E}^f=\frac{1}{2}{k}_s{x}^2$

the downward potential energy of the jumper $P{E}^f$that is the half of the product of the stiffness of the spring$k_s$ and square of the displacement x.

According to the law of energy conservation, the equation of the energy conservation can be expressed as:

$E_1=E_2$

Here,$E_1$ is the energy at the top and$E_2$ is the energy at the bottom.

The above equation can also be expressed as:

$K.E.+P{E}_g+P{E}_2=K.E^f+P{E}_g^f+P{E}_2^f$

Here, K.E. is the kinetic energy of the jumper at the top,$P{E}_g$is the potential energy of the jumper at the top due to gravity,$P{E}_s$is the potential energy of one cord ,$K.E^f.$is the kinetic energy of the jumper at the top,$P{E}_g^f$ is the potential energy of the jumper at the top due to gravity and$P{E}_s^f$ is the potential energy of another cord

As the potential energy at the top and the bottom due to gravity are same, hence, the above equation can be expressed as:

$P{E}_g+PE=P{E}_g^f+P{E}^f$

According to the Hooke’s law and the energy conservation, the above equation can be demonstrated as:

$mgh-\frac{1}{2}k{x}^2=mgx+\frac{1}{2}k{x}^2$

Substituting all the values in the above equation.

$$\begin{gathered} 80\mathrm{kg}\times 9.81\mathrm{m}/{\mathrm{s}}^2\times \left(10\mathrm{m}-\left(-24\mathrm{m}\right)\right)=\mathrm{k}{\left(-24\mathrm{m}\right)}^2 \\ \mathrm{k}=\frac{784.8\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2\times 34\mathrm{m}}{576{\mathrm{m}}^2} \end{gathered}$$

$=46.325\mathrm{kg}/{\mathrm{s}}^2$

Thus, the spring stiffness for each cord is $46.325\mathrm{kg}/{\mathrm{s}}^2$.

According to the Hooke’s law, the equation for the maximum tension is expressed as:

$F_s=-kx$

Here,$F_s$ is the maximum tension,k is stiffness of the cords and x is the maximum displacement of the cords

Substituting the values in the above equation.

$$\begin{gathered} F_s=-(46.325\mathrm{kg}/{\mathrm{s}}^2\times -24\mathrm{m}) \\ =1111.8\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2 \\ =1111.8\mathrm{N} \end{gathered}$$

Thus, the maximum tension that each cord must support without breaking is 11118.N

From the Newton’s second law, the equation of maximum acceleration of the jumper is expressed as:

$$\begin{gathered} F=ma \\ a=\frac{F}{m} \end{gathered}$$

Here,F is the maximum tension of the cords,m is the mass of the jumper and aa is the maximum acceleration of the jumper

It is known that,

$$\begin{gathered} F_s=ma+mg \\ {F}_s=F+mg \\ F=F_s-mg \end{gathered}$$

As there are two cords, thus the value of the force is $F=2F_s-mg$.

Using the value of the force in the above equation,

$a=\frac{2F_s-mg}{m}$

Substituting the values in the above equation.

$$\begin{gathered} a=\frac{2\times 1111.8\mathrm{N}-80\mathrm{kg}\times \mathrm{g}}{80\mathrm{kg}} \\ =27.795\mathrm{N}/\mathrm{kg}-g \\ =17.985\mathrm{N}/\mathrm{kg}\times \frac{1\mathrm{kg}·{\mathrm{ms}}^2/\mathrm{kg}}{1\mathrm{N}/\mathrm{kg}}-g \\ =27.795\mathrm{m}/{\mathrm{s}}^2-g \end{gathered}$$

Hence, further simplified as:

$$\begin{gathered} a=27.795\mathrm{m}/{\mathrm{s}}^2-g \\ =2.83\times \left(9.8\mathrm{m}/{\mathrm{s}}^2\right)-g \\ =2.83g-g \\ =1.83g \end{gathered}$$

Thus, the maximum acceleration that the jumper experiences is 1.83g and the direction of this acceleration is upward as the value is positive.

The air resistance is neglected and Hooke’s law is being obeyed for the elastic cord. The initial and the net change in the kinetic energy has been gathered with the help of these approximations. Moreover, the Hooke’s law also helped in identifying the maximum tension.

Thus, the air resistance is neglected and the Hooke’s law is being obeyed.