A 53 kg bungee jumper jumps from a bridge. She is tied to a bungee cord whose unstretched length is 13 m, and falls a total of 35 m. Calculate the spring stiffness constant k of the bungee cord, assuming Hooke's law applies.

Let X be the deflection from the zero-stretch length (13).

When motion is considered, max spring potential energy (at max stretch) equals the gravitational potential energy loss at maximum extension.

(1/2)k Xmax^2 = M g *35 m

Xmax = 22 m

k = 2 M g *35m/(22)^2
= 75.1 Newton/meter

Its elastic potential energy

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To calculate the spring stiffness constant (k) of the bungee cord, we need to use Hooke's law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.

In this scenario, the displacement is the difference between the unstretched length of the bungee cord and the total distance fallen by the jumper:

Displacement (x) = 35 m - 13 m = 22 m

Now, let's calculate the force exerted by the spring (F) on the bungee jumper. The force can be calculated using Hooke's law:

F = k * x

Where:
F is the force exerted by the spring
k is the spring stiffness constant
x is the displacement

Given that the bungee jumper weighs 53 kg, we can calculate the force due to gravity (mg) acting on her:

Force due to gravity (mg) = mass (m) * acceleration due to gravity (g)

Where:
m is the mass of the bungee jumper (53 kg)
g is the acceleration due to gravity (9.8 m/s^2)

Force due to gravity (mg) = 53 kg * 9.8 m/s^2 = 519.4 N

Since the bungee cord counteracts the force due to gravity, the force exerted by the spring must be equal to the force due to gravity:

F = 519.4 N

Now, we can plug in the known values into Hooke's law equation and solve for the spring stiffness constant:

519.4 N = k * 22 m

Divide both sides by 22 m:

k = 519.4 N / 22 m

Calculating the value:

k ≈ 23.61 N/m

Therefore, the spring stiffness constant (k) of the bungee cord is approximately 23.61 N/m.