Mechanics (bungy jumping)

用戶81429

2007-06-10 7:35 am

A man with mass 70kg is is bungy jumping from 200m above a river. Length of cord used is 50m

1 Calculate the velocity when the cord is just stretched

2 The man stops when he just reached river, calculate the extension and the energy stored by the elastic cord

3 Given that energy stored in the cord =(1/2)kx^2, k is the force constant of cord and x the extension of cord (force constant=force per unit extension) Find:
i) k
ii) acc. of the man the moment when he reaches water

回答 (1)

天同

2007-06-10 8:47 am

✔ 最佳答案

1. Using conservation of mechanical energy,
(1/2)(70)v^2 = (70)g.(50)
where v is the velocity when the cord is just stretched and g is the acceleration due to gravity
solving for v
2. By conservation of energy, the energy stored in the cord is just equal to the total potential energy loss of the man.
thus, energy stored by the cord = 70g(200) J
The extension of the cord = (200-50) m = 150 m
3. 70g(200) = (1/2)k(150)^2
solve for k
The net force acting on the man = kx - mg
thus (kx-mg) = ma
where a is the acceleration
a = (kx-mg)/m
since k, x, m and g are known, calculate a

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