Question

a ride at an amusement park attaches people to a bungee cord, pulls them straight down to the ground, and then releases them into the air. when they are pulled to the ground, the bungee cord (which has a stiffness constant of 35 n/m) is stretched 100 m beyond its unloaded length. what restraining force is required to hold a man with a mass of 90 kg to the ground just before he is released? (recall that ( g = 9.8 mathrm{~m} / mathrm{s}^{2} ))
a. 3500 n
b. 2830 n
c. 2618 n
d. 2670 n

Explanation:

Step1: Calculate the force exerted by the bungee cord

Using Hooke's Law \(F = kx\), where \(k = 35\space N/m\) and \(x=100\space m\).
\(F_{bungee}=35\times100 = 3500\space N\)

Step2: Calculate the weight of the man

Using \(W = mg\), where \(m = 90\space kg\) and \(g = 9.8\space m/s^{2}\).
\(W=90\times9.8=882\space N\)

Step3: Find the restraining force

At equilibrium before release, \(F_{bungee}=W + F_{restraining}\).
So, \(F_{restraining}=F_{bungee}-W\)
Substitute the values: \(F_{restraining}=3500 - 882=2618\space N\)

Answer:

C. 2618 N