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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 à 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 80 kg to the ground just before
he is released? (Recall that g = 9.8 m/s²)
100 m
Force of bungee
Weight+ Restraining Force

Respuesta :

onyebuchinnaji

The restraining force is required to hold a man with a mass of 80 kg to the ground just before he is released is 2,716 N.

Force required to hold down the man

The total force required to hold down the man of the given weight is calculated as follows;

Force of bungee = Weight + Restraining Force

F = W + Rf

Rf = F - W

F = (35 x 100) - (80 x 9.8)

F = 2,716 N

Thus, the restraining force is required to hold a man with a mass of 80 kg to the ground just before he is released is 2,716 N.

Learn more about restraining force here: https://brainly.com/question/12970081

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