A 100kg man steps off a roof and free-falls,he is attached to a 50m long, with k=1000N\cm.?

Question

what will be the maximum stretch of the line at the instant he comes to rest, assuming it remains Hookean.

Answer 1

The number is 9

Answer 2

The line's 50 m long, so his gravitational potential energy is 50m * g * 100kg = 50,000 J.

He'll have 50,000 J of kinetic energy when he hits the spring.

The potential energy stored in a spring is given by:

1/2 * k * x^2.

Here, k = 100,000 N/m. I converted it to meters.

So: 1/2 k x^2 = 50,000 J

k x^2 = 100,000 J

100,000 N/m * x^2 = 100,000 J (or N-m)

x ^ 2 = 1 m^2

So he stretches it 1 m or 100 cm.

Oh, I have no earthly idea what "hookean" means.

[EDIT]
I approximated g at 10m/s^2. Technically it's actually 9.8m/s^2, which would yield a slightly lower number, but the way all the numbers work out pretty I'd venture a guess that the original problem intended for you to approximate g at 10m/s^2.

Substituting the exact value of g is relatively straightforward, but won't change the answer until the 2nd decimal place.
[/EDIT]

Answer 3

who cares, the guy will still be dead. If the decelleration from the rope stopping him doesn't kill him, either hitting the ground or the side of the building at 100MPH will