Part A
(Photo courtesy Wikimedia Commons, uploaded by user Che010.)
Bungee cords designed to U.S. Military specifications (DoD standard MIL-C-5651D, available at http://dodssp.daps.dla.mil
) are characterized by a force constant times unstretched length in the range kL ~ 800-1500 N. Jumpers using these cords intertwine three to five cords to make a thick rope that is strong enough to withstand the forces of the jump. Suppose that you are designing a bungee jump off of a bridge that is 30.0 m above the surface of a river running below. To get an idea for the maximum cord length, calculate the unstretched length of cord with kL = 4000 N (a system of five 800 N cords) that will result in a 118 kg jumper who leaves the bridge from rest ending their jump 2.0 m above the water's surface. (The 2.0 m builds in a safety margin for the height of the jumper, so you can neglect the height in your calculation.) Since you are finding a maximum cord length, ignore any losses due to air resistance or dissipation in the cord. Ignore the mass of the rope.
System: The jumper (treated as a point particle) plus the earth and the bungee cord. The system constituents interact via gravity, which contributes [gravitational potential energy], and via the restoring force of the cord, which contributes [elastic potential energy]. External influences are assumed negligible.
Model: [Constant Mechanical Energy].
Approach:
Shown above is an initial-state final-state diagram for this situation, along with corresponding energy bar graphs. As indicated in the picture, we have chosen the zero point of the height to be the river's surface. With these pictures in mind, we can set up the Law of Change for our model:
\begin
[ E_
= mgh_
= E_
= mgh_
+ \frac
k x_
^
] \end
Bungee cords provide a restoring force when stretched, but offer no resistance when "compressed", since they fold like an ordinary rope. Thus, the initial spring energy is zero in this case.
This equation cannot be solved without further constraints, since we do not know k. The extra constraint that we have is given by the fact that the jumper has fallen a total of 28.0 m (descending from 30.0 m above the water down to 2.0 m above the water). This distance must be covered by the stretched cord. This gives us the contraint:
\begin
[ h_
- h_
= L + x_
] \end
Solving this constraint for xf and substituting into the energy equation gives:
\begin
[ -2mg(\Delta h) = k((\Delta h)^
+ 2 L \Delta h + L^
) ] \end
Where we are using Δh = hf - hi to simplify the expression.
We still have two unknowns, but we can resolve this by using the fact that kL is a constant for the rope. Thus, if we multiply both sides by L, we have:
\begin
[ -2mg(\Delta h)L = C((\Delta h)^
+ 2L \Delta h + L^
) ] \end
where we have replaced the quantity kL by C (= 800 N) for clarity. With this substitution, it can be seen that we have a quadratic equation in L which can be solved to find:
\begin
[ L = \frac{-2(C+mg)\Delta h \pm \sqrt{4(C+mg)^
(\Delta h)^
-4C^
(\Delta h)^
}}
= 13.3 \:
\;
\;58.9 \:
] \end
It is clear that the appropriate choice is 13.3 m.
Part B
Suppose a 118 kg jumper foiled your calculations by leaping upward from the bridge with an initial speed of 2.25 m/s. Will the jumper hit the water, assuming the 13.3 m cord with kL = 4000 N that we found in Part A?