91影视

The given data is listed below as,

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

Hooke鈥檚 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鈥檚 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鈥檚 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,localid="1658467390824" $v_B$is the velocity of the body andlocalid="1658467399985" $k{s}_B$is the upward force. In the diagram C and D,localid="1658467408351" $k{s}_C$andlocalid="1658467417742" $k{s}_D$is the upward force andlocalid="1658467431483" $v_C$andlocalid="1658467451458" $v_D$are the velocity of the body. In the diagram Elocalid="1658467461174" $v_E$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鈥檚 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:

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

Here,k is the spring constant, $S_f$is the displacement, m is the mass,g is the gravitational acceleration,$h_i$ is the initial and$h_f$ is the final height and $v_f$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鈥檚 speed at this instant is 0.

The value of$d\stackrel{-}{p}/dt$ can be written as mdV/dt as the momentum is the product of the mass and the velocity of the jumper The jumper鈥檚 momentum is changing due to the reason that the velocity of the jumper is continuously changing. So, the value of $d\stackrel{-}{p}/dt$is not 0.

Thus, the jumper鈥檚 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 the cord length h.

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:

$P{E}_g^{,}=mgx$

The potential energy at the bottom$P{E}_g^{,}$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:

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

the downward potential energy of the jumper$P{E}^{,}$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:

$K.E.+P{E}_g+P{E}_s=K.E^{,}.+P{E}_g^{,}+P{E}_s^{,}$

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' ,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 and$P{E}_s^{,}$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:

$P{E}_g+PE=P{E}_g^{,}+PE\text{'}$

According to the Hooke鈥檚 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.

$$\begin{gathered} 80\mathrm{kg}脳9.81\mathrm{m}/{\mathrm{s}}^{}脳\left(10\mathrm{m}-\left(-24\mathrm{m}\right)\right)=\mathrm{k}{\left(-24\mathrm{m}\right)}^2 \\ \mathrm{k}=\frac{784.8\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2脳34\mathrm{m}}{576{\mathrm{m}}^2} \\ =46.325\mathrm{kg}/{\mathrm{s}}^2 \end{gathered}$$

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

According to the Hooke鈥檚 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.

$$\begin{gathered} F_s=-\left(46.325\mathrm{kg}/{\mathrm{s}}^2脳-24\mathrm{m}\right) \\ =1111.8\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \\ =1111.8\mathrm{N} \end{gathered}$$

Thus, the maximum tension that each cord must support without breaking is 1111.8 N.

From the Newton鈥檚 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 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.

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

Hence, further simplified as:

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

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

The air resistance is neglected and Hooke鈥檚 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鈥檚 law also helped in identifying the maximum tension.

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