Question

Claudia goes bungee jumping. She has a mass of 51 kg.

a. The bungee cord will stretch 5 m when 200 N of force is applied to it. What is the spring constant of the bungee cord?

b. After her jump, Claudia hangs from the cord motionless. How stretched is the bungee cord past its natural length?

c. The maximum force the bungee cord can handle without breaking is 15,000 N. How far would you have to stretch it to get it to break?

Answer 1

a) According to Hooke's law:

k=F/m=40 N/m.

b) x=mg/k=12.5 m

c) x=F.max/k=375 m.

Answer 2

Here Bungee will behave like an ideal spring

PART A)

so we know that spring force is given as

`F = kx`

here we know that

F = 200 N

x = 5 m

now from above equation

`200 = k (5)`

`k = 40 N/m`

PART b)

As we know that mass of the Claudia = 51 kg

now its force due to weight is given as

`F = mg`

`F = 51(9.8) = 499.8 N`

now from the above formula of spring

`F = kx`

`499.8 = 40(x)`

`x = 12.5 m`

so bungee will be stretch by 12.5 m distance

Part C)

Maximum force on the bungee is given as

`F = kx`

`F = 15000 N`

`15000 = 40 x`

`x = 375 m`

so it will stretch by 375 m distance