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246 CHAPTER 8 Conservation of Energy

Some fanatics, in search of dangerous thrills, jump off high
EXAMPLE 4
bridges or towers with bungee cords (long rubber cords) tied
to their ankles (Fig. 8.11). Consider a jumper of mass 70 kg, with a 9.0-m cord
tied to his ankles. When stretched, this cord may be treated as a spring, of spring
constant 150 N/m. Plot the potential-energy curve for the jumper, and from this
curve estimate the turning point of the motion, that is, the point at which the
stretched cord stops the downward motion of the jumper.

SOLUTION: It is convenient to arrange the x axis vertically upward, with the origin
at the point where the rubber cord becomes taut, that is, 9.0 m below the jump-off
point (see Fig. 8.12a). The potential-energy function then consists of two pieces.
For x 0, the rubber cord is slack, and the potential energy is purely gravitational:
U mgx for x 0
For x 0, the rubber cord is stretched, and the potential energy is a sum of gravi-
tational and elastic potential energies:
1
2
U mgx kx for x 0
2
With the numbers specified for this problem,
FIGURE 8.11 Bungee jumping. 2
U 70 kg 9.81 m/s x
687x for x 0 (8.24)

and
2 1 2
U 70 kg 9.81 m/s x 150 N/m x
2
2
687x 75x for x 0 (8.25)
where x is in meters and U in joules. Figure 8.12b gives the plot of the curve of
potential energy, according to Eqs. (8.24) and (8.25).
At the jump-off point x 9.0 m, the potential energy is U 687x 687
9.0 J 6180 J.The red line in Fig. 8.12b indicates this energy level.The left inter-
section of the red line with the curve indicates the turning point at the lower end
of the motion. By inspection of the plot, we see that this turning point is at x
15 m.Thus, the jumper falls a total distance of 9.0 m 15 m 24 m before his
downward motion is arrested.
We can accurately calculate the position of the lower turning point (x 0)
by equating the potential energy at that point with the initial potential energy:

2
687x 75x 6180 J
2
This provides a quadratic equation of the form ax bx c 0:
2
75x 687x 6180 0

2
This has the standard solution x ( b ; 2b 4ac ) 2a, or
2
687; 3 (687) 4 75 6180
x
2 75
14.7 m 15 m
in agreement with our graphical result. Here we have chosen the negative solu-
tion, since we are solving for x at the lower turning point using the form (8.25), which
is valid only for x 0.