A bungee jumper of mass 75 kg uses a bungee cord with a relaxedlength of 50 m and jumps from a bridge that is 170 m above the ground. What elastic force constant is necessary for the bungee cord in order for him to jump safely.
I have no idea what to do or even start this problem
To solve this problem, we need to use the concepts of gravitational potential energy and elastic potential energy. Let's break down the problem into smaller steps:
Step 1: Determine the distance the bungee jumper falls
The bungee jumper jumps from a bridge that is 170 m above the ground. Upon jumping, the bungee cord will stretch to its maximum length and then pull the jumper back up. To find the distance the bungee jumper falls, we subtract the relaxed length of the bungee cord from the height of the bridge:
Distance fallen = Bridge height - Relaxed length of cord
= 170 m - 50 m
= 120 m
Step 2: Calculate the gravitational potential energy
The gravitational potential energy (GPE) is given by the formula:
GPE = m * g * h
where
m = mass of the bungee jumper (75 kg),
g = acceleration due to gravity (9.8 m/s^2),
h = height fallen (120 m).
Plugging in the values, we get:
GPE = 75 kg * 9.8 m/s^2 * 120 m
= 88,200 J
Step 3: Calculate the elastic potential energy
The elastic potential energy (EPE) is given by the formula:
EPE = (1/2) * k * x^2
where
k = elastic force constant of the bungee cord (unknown),
x = distance stretched (120 m).
Since the bungee cord starts from its relaxed length, its maximum stretch is equal to the distance the bungee jumper falls. So, x = 120 m.
Plugging in the values, we get:
EPE = (1/2) * k * (120 m)^2
= 7200 k
Step 4: Equate the gravitational potential energy to the elastic potential energy
Since the bungee jumper is safe, the elastic potential energy must be equal to or greater than the gravitational potential energy. Therefore:
EPE ≥ GPE
7200 k ≥ 88,200
Solving for k:
k ≥ (88,200) / (7200)
k ≥ 12.25
Therefore, the elastic force constant necessary for the bungee cord to ensure a safe jump is equal to or greater than 12.25 N/m.