a bungee jumper of mass 80kg jumps from 20m, bungee cord has an original length=4m, end of cord in 2m from the bottom when jumper is momentarily at rest at the bottom. find k value.
why can't you use f=kx to solve, where f=mg and x=14?
because the stretch stops when the kinetic energy has all been stored in the cord, which is 1/2 kx^2
To calculate the spring constant (k) in this scenario, we need to determine the effective length of the cord at the instant the jumper is momentarily at rest at the bottom. This effective length accounts for the stretching of the cord beyond its original length by the weight of the jumper.
Using the given information:
- The original length of the cord (L) is 4m
- The bungee jumper jumps from a height (h) of 20m
- The end of the cord is 2m from the bottom when the jumper is momentarily at rest at the bottom
- The mass of the jumper (m) is 80kg
To calculate the effective length, we subtract the distance from the bottom where the jumper is momentarily at rest from the total length of the cord:
Effective length = L - distance from bottom where the jumper is momentarily at rest
Effective length = 4m - 2m
Effective length = 2m
Now we can use Hooke's Law to calculate the spring constant (k) using the formula:
F = k * x
Where:
- F is the force applied to the spring (in this case, the weight of the jumper = mg)
- k is the spring constant
- x is the displacement from the equilibrium position (in this case, the effective length of the cord = 2m)
By substituting the values:
mg = k * x
80kg * 9.8 m/s^2 = k * 2m
784 N = 2k
k = 784 N / 2m
k = 392 N/m
Therefore, the value of the spring constant (k) is 392 N/m.
Now, regarding your question about using f = kx, where f = mg and x = 14:
The equation f = kx represents Hooke's Law, where f is the force exerted by a spring, k is the spring constant, and x is the displacement from the equilibrium position. It is not applicable in this case because x = 14m does not represent the displacement of the cord from its equilibrium position when the jumper is momentarily at rest at the bottom. The effective length of the cord at that point is 2m, not 14m.