Share
In the dangerous “sport” of bungee-jumping, a daring student jumps from a balloon with a specially designed elastic cord attached to his wai
Question
In the dangerous “sport” of bungee-jumping, a daring student jumps from a balloon with a specially designed elastic cord attached to his waist. The unstretched length of the cord is 23.4 m, the student weighs 818 N, and the balloon is 31.3 m above the surface of a river below. Calculate the required force constant of the cord if the student is to stop safely 2.74 m above the river. Answer in units of N/m.
in progress
0
Physics
3 years
2021-09-02T13:18:52+00:00
2021-09-02T13:18:52+00:00 1 Answers
12 views
0
Answers ( )
Answer:
k = 1755 N/m
Explanation:
Given:
– The length of the cord L = 23.4 m
– Weight of the student W = 818 N
– The elevation of balloon H = 31.3 m
Find:
Calculate the required force constant of the cord if the student is to stop safely 2.74 m above the river.
Solution:
– We know the potential energy of the student changes by
ΔP.E = m*g*( H – 2.74 )
mg*(31.3 – 2.74) = 818*28.56 = 23362.08 J
– When he stops at 2.74 m above ground his KE = 0 so ALL his lost potential energy must be stored in the extended or stretched bungee cord.
– He falls 23.4 m before the bungee cord starts to stretch. That means it doesn’t start stretching until he is 31.3 – 23.4 = 7.9 m above the ground.
– It has to stop stretching at 2.74 m above the ground so the
total stretch = 7.9 – 2.74 = 5.16 m
– Therefore his PE from 31.3 m to 2.74 m is stored in a 5.16 m stretch of the bungee cord.
½kx² = ΔP.E
k = 2*ΔP.E / x^2
k = 2*23362.08 / 5.16^2
k = 1755 N/m