(II) Chris jumps off a bridge with a bungee cord (a heavy stretchable cord) tied around his ankle, Fig. 6–42. He falls for 15 m before the bungee cord begins to stretch. Chris’s mass is 75 kg and we assume the cord obeys Hooke’s law,with If we neglect air resistance, estimate what distance dbelow the bridge Chris’s foot will be before coming to a stop. Ignore the mass of the cord (not realistic, however) and treat Chris as a particle.

FIGURE 6–42Problem 41. (a) Bungeejumper about to jump. (b) Bungee cord at itsunstretched length.(c) Maximum stretchof cord.

Chris jumped from the bridge and tied a bungee cord to his ankle. The bungee cord behaves like a spring.

The conservation of energy principle needs to be applied here. During the jump, the loss in the gravitational potential energy due to the vertical distance is stored in the form of the cord’s elastic potential energy.

The total distance covered in the downward motion is the sum of the length of the cord and the distance stretched by the cord below the bridge.

The given data can be listed below as:

• The length of the bungee cord is\(h = 15{\rm{ m}}\).
• The mass of Chris is\(m = 75{\rm{ kg}}\).
• The stiffness constant of the cord is\(k = 55{\rm{ N/m}}\).
• The acceleration due to gravity is\(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).

From the conservation of energy principle, the potential energy loss during the downward motion is equal to the gain in the elastic potential energy,and thiscan be expressed as:

\(\begin{aligned}P.E &= E.P.E\\mgd &= \frac{1}{2}k{\left( {\Delta y} \right)^2}\\mg\left( {h + \Delta y} \right) &= \frac{1}{2}k{\left( {\Delta y} \right)^2}\\mgh + mg\Delta y &= \frac{1}{2}k{\left( {\Delta y} \right)^2}\end{aligned}\)

Here, \(E.P.E\) is the elastic potential energy of the cord, \(P.E\) is the potential energy of Chris, \(\Delta y\) is the stretched length of the cord, and d is the total distance below the bridge before he comes to rest.

Substitute the values in the above expression.

\(\begin{aligned}75{\rm{ kg}} \times 9.81{\rm{ m/}}{{\rm{s}}^2} \times 15{\rm{ m}} + 75{\rm{ kg}} \times 9.81{\rm{ m/}}{{\rm{s}}^2} \times \Delta y &= \frac{1}{2}\left( {55{\rm{ N/m}}\left( {\frac{{1{\rm{ kg/}}{{\rm{s}}^2}}}{{{\rm{N/m}}}}} \right)} \right) \times {\left( {\Delta y} \right)^2}\\11036.25 + 735.75\Delta y &= 27.5{\left( {\Delta y} \right)^2}\\27.5{\left( {\Delta y} \right)^2} - 735.75\Delta y - 11036.25 &= 0\\\Delta y &= 37.466{\rm{ m}}\end{aligned}\)

The total distance below the bridge and Chris’s foot before he comes to rest position can be expressed as:

\(d = h + \Delta y\)

Here, d is the total distance.

Substitute the values in the above expression.

\(\begin{aligned}d &= 15{\rm{ m}} + 37.466{\rm{ m}}\\ &\approx 52.47{\rm{ m}}\end{aligned}\)

Thus, the total distance below the bridge and Chris’s foot before coming to rest is 52.47 m.