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 100 kg to the ground just before he is released? (Recall that g = 9.8 m/s²) Force of bungee 100 m Weight + Restraining Force

A. 2670 N
B. 2430 N
C. 2520 N
D. 2830 N

Answer 1

The restraining force required to hold the man to the ground just before he is released is 4480 N.

Step-by-step explanation:

To find the restraining force required to hold the man to the ground, we need to consider the weight of the man and the force exerted by the bungee cord. The weight of the man can be calculated using the formula weight = mass x gravitational acceleration. In this case, the mass is given as 100 kg and the gravitational acceleration is 9.8 m/s². So the weight of the man is 100 kg x 9.8 m/s² = 980 N.

The force exerted by the bungee cord is equal to its stiffness constant multiplied by the amount it is stretched. In this case, the stiffness constant is 35 N/m and the bungee cord is stretched 100 m beyond its unloaded length. So the force exerted by the bungee cord is 35 N/m x 100 m = 3500 N.

The restraining force required to hold the man to the ground is the sum of the weight and the force exerted by the bungee cord. Therefore, the restraining force is 980 N + 3500 N = 4480 N.