http://www.forrestrace.com/breaker-grozier-2/
How much work is required to stretch the elastic cord?
An elastic cord (elastic constant, k) of unstretched length L is tied to two posts at (0, 0) and (L, 0). A pair of pliers ( http://upload.wikimedia.org/wikipedia/commons/b/bc/Breaker-grozier-pliers.jpg ) is used to grab the cord at its midpoint (L/2, 0) and drag it to point (L, L/2).
Find an expression for the amount of work required to accomplish this.
Can you also generalize the result? Let’s say you grab the cord at (a, 0) and drag it to (0, a), where 0 < a < L.
Sorry for the generalized case, you're grabbing the cord at (a, 0) and draging it to (L, a).
Yes I keep saying the wrong thing. The general case is (a,0) dragging to (0,a) or (L, L-a).
Let me reiterate that we're GRABBING the cord with the pliers. We're NOT letting the cord SLIDE over the pliers.
So are you guys saying that it makes no difference whether we grab the cord or whether we let it slide over the pliers?
It definitely matters whether you slide or grab. If you slide, you are working with one spring. If you grab, you have two separate springs. If each spring is half as long (ie, you grab in the middle), then each is twice as stiff. The force is linear, but the energy stored in the string is quadratic with respect to the length of stretch. So if one stretches more and the other less, that matters. Also, if one of the segments gets shorter, that fouls up the calculations further. Also, at some point, Hooke’s law won’t hold for the cord (which doesn’t store compressive energy well in any case).
Mad’s solution is basically good, but I think he may be off by a factor of two. The half cord has spring constant 2k.
How to cut glass – small thin strips
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