Conservation of mechanical energy of bungee jumper

[joej]

A 60kg bungee jumper jumps from a bridge. He is tied to a 12m-long bungee cord and falls a total of 31m
a) calculate the spring constant of the bungee cord
b) calculate the maximum acceleration experienced by the jumper

okay I'm stuck on b) I got a0 calculated without any problems at all, it will be 101N/m the problem with b) is that I initially thought... how would it be possible to have an acceleration of more then 9.8m/s^2 while falling... it isn't! but I never took into consideration the acceleration while going back up and now whatever I do I alwasy come back to the 9.8m/s^2 answer, I need someon to point me in the right direction so I can get off of this damn idea of 9.8m/s^2

 

[arildno]

When the cord is stretched maximally, you will experience maximal upwards acceleration.
The upwards acceleration is found by subtracting mg (downwards force) from the "spring force" at maximal stretch; and then divide by "m", the mass of the jumper.
Compare then that value of maximal upwards acceleration with 9.8 (maximal downwards acceleration)

 

[wais]

a) To calculate the spring constant of the bungee cord, we can use the conservation of mechanical energy equation:

E_initial = E_final

Where E_initial is the initial mechanical energy (potential energy + kinetic energy) and E_final is the final mechanical energy (potential energy + kinetic energy).

We know that the bungee jumper has a mass of 60kg and falls a total of 31m. We also know that the bungee cord is 12m long.

E_initial = mgh = (60kg)(9.8m/s^2)(31m) = 18,108 J

To find the final mechanical energy, we need to consider the potential energy at the lowest point (when the bungee cord is fully extended) and the kinetic energy at that point.

Potential energy at lowest point = mgh = (60kg)(9.8m/s^2)(12m) = 7,056 J

Kinetic energy at lowest point = (1/2)mv^2

To find the velocity at the lowest point, we can use the conservation of energy equation again:

E_initial = E_final

18,108 J = 7,056 J + (1/2)(60kg)v^2

Solving for v, we get v = 13.42 m/s

Therefore, the kinetic energy at the lowest point is (1/2)(60kg)(13.42 m/s)^2 = 5,085 J

The final mechanical energy is then 7,056 J + 5,085 J = 12,141 J

Now, we can use this to find the spring constant:

E_final = (1/2)kx^2

Where x is the extension of the bungee cord. At the lowest point, the bungee cord is fully extended, so x = 12m.

12,141 J = (1/2)k(12m)^2

Solving for k, we get k = 84.38 N/m

b) The maximum acceleration experienced by the jumper can be found using Newton's second law:

F = ma

Where F is the force acting on the jumper, m is the mass of the jumper, and a is the acceleration.

At the lowest point, the only force acting on the jumper is the tension in the bungee cord. This