A 72 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 38 m. Calculate the spring stiffness constant "k" of the bungee cord, assuming Hooke's law applies. Calculate the maximum acceleration she experiences.
mgh=1/2 k x^2
72(9.8)*38=1/2 k (38-13)^2
solve for k
Maximum acceleration? as she begins to fall, it is -9.8m/s^2
Now at the bottom, she is accelerating upward.
Forcebottom= -mg+k(38-13)
and Force= ma
solve for a.
To calculate the spring stiffness constant "k" of the bungee cord, we can use Hooke's law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
Hooke's law equation: F = -kx
In this case, the force acting on the bungee jumper is her weight, which is given by the formula: F = m * g, where m is the mass of the jumper and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Given:
Mass of the bungee jumper (m) = 72 kg
Displacement of the bungee cord (x) = 38 m
Substituting the values into Hooke's law equation:
m * g = -kx
Rearranging the equation to solve for "k":
k = -(m * g) / x
Plugging in the values:
k = - (72 kg * 9.8 m/s^2) / 38 m
Calculating the spring stiffness constant "k":
k = - 705.6 N/m
So the spring stiffness constant "k" of the bungee cord is approximately -705.6 N/m.
Now, let's calculate the maximum acceleration experienced by the bungee jumper.
The maximum acceleration occurs when the bungee cord is stretched to its maximum length. At this point, the force exerted by the bungee cord is equal to the weight of the jumper, which can be calculated using the formula: F = m * g.
Given:
Mass of the bungee jumper (m) = 72 kg
Acceleration due to gravity (g) = 9.8 m/s^2
Maximum acceleration (a) = F / m
Substituting the values:
a = (72 kg * 9.8 m/s^2) / 72 kg
Calculating the maximum acceleration:
a = 9.8 m/s^2
So, the bungee jumper experiences a maximum acceleration of 9.8 m/s^2.