Simple Harmonic Motion: Fire Net Stretch Calculations

[FearTheHump]

Simple harmonic motion! :(

Homework Statement

A 65-kg person jumps from a window to a fire net 18m below, which stretches the net 1.1m. Assume that the net behaves like a simple spring, and (a) calculate how much it would stretch if the same person were lying in it. (b) How much would it stretch if the person jumped from 35m?

Homework Equations

No idea. All of the equations I have seem to reference acceleration, or velocity, or force...etc. I can't figure out what to use for this.

The Attempt at a Solution

I have no idea where to start with this question. What confuses me the most is the whole "falling 18 metres" or..."falling 35 metres". Now, I have got massive holes in my physics knowledge (slacked off a lot last year, I will admit), but from what I understand, f=ma, which means that the resulting force will be the same for both drops, right? Well, logically, no, but from my limited physics knowledge, apparently.

Please help me not fail physics once again! :(

 

[Doc Al]

Presumably they want the maximum stretch of the net. Hint: Use conservation of energy. What's the formula for calculating the energy stored in a stretched spring?

 

[FearTheHump]

EPE = 1/2kx^2 right?
So for the 18m situation where i know x, would I use that to find k, then sub that into get x in the first question?

That would make sense to me, but usually things that make sense to me in Physics are wrong.

But anyway...
GPE(o) = KE (which is zero) + EPE(f)
mgh = 1/2kx^2
65 X 9.8 X 18 = 1/2 X k X 1.21
11466 = 0.605k
k = 18,952?

a) Still...no idea?

b) GPE(o) = KE (zero again) + EPE(f)
mgh = 1/2kx^2
65x9.8x35 = 1/2 X 18952 X x^2
22925 = 9476x^2
x = (square root of)2.42m
= 1.56m

This sounds about right, but I still have my doubts.PS - I've used capital X for multiplication, and lower case x for the variable.

 

[Doc Al]

EPE = 1/2kx^2 right?
So for the 18m situation where i know x, would I use that to find k, then sub that into get x in the first question?

Right.

But anyway...
GPE(o) = KE (which is zero) + EPE(f)
mgh = 1/2kx^2
65 X 9.8 X 18 = 1/2 X k X 1.21
11466 = 0.605k
k = 18,952?

Almost right. Hint: Measure gravitational PE from the lowest point. (The person doesn't just fall 18 m, but 18 + x.)

a) Still...no idea?

Once you've got k, use Hooke's law.

b) GPE(o) = KE (zero again) + EPE(f)
mgh = 1/2kx^2
65x9.8x35 = 1/2 X 18952 X x
22925 = 9476x
x = 2.42m

You'll need to redo this, with the same caveat as above. (And careful with x^2 versus x.)

 

[FearTheHump]

You'll need to redo this, with the same caveat as above. (And careful with x^2 versus x.)

Thanks, but...how am I to know how much to add to the height in (b)?
Since the person drops from a higher position, it will be a larger number, right? I don't know what that number is, though, because that's what the whole equation is trying to solve.
:/

 

[Doc Al]

Thanks, but...how am I to know how much to add to the height in (b)?
Since the person drops from a higher position, it will be a larger number, right? I don't know what that number is, though, because that's what the whole equation is trying to solve.

Exactly. Since you don't know the amount that the net stretches in (b), call the unknown stretch x, set up your equations, and solve for it.

 

[FearTheHump]

Ahhhh, a quadratic. Thanks! :D