Why is the formula for elastic potential energy in the 1st part not 0.5Fx ?
Good question. See the exchange below, where James asked essentially the same question two years ago.
The simple answer is that the PE does equal Fx/2 if the bow works like a perfect spring, in which the force is proportional to the distance that the arrow is pulled back. But that is not how bows work. In fact the perfect bow would have a near constant draw force as it is extended (see this high quality bow for an example http://archeryreport.com/wp-content/uploads/2011/02/envy_draw_force_curve.jpg). So I have assumed that the bow in the question is some kind of ‘ideal bow’, in which case the PE equals Fx.
There’s no right answer here, I’m afraid, because it is really a silly question. My guess is that the person who set the question was probably a physicist rather than an engineer, naively assumed that a bow works like a perfect spring, and was expecting the answer that you have gone for. But I’ve gone for a different answer to bring out the fact that the draw force curve of the bow hasn’t been described.
Probably the perfect answer would be to state explicitly that the PE is the integral of force with respect to distance but we can’t work out the exact PE here because the force/distance curve hasn’t been given, so we are going to assume … (then put in whatever assumption you want to make, either that the bow obeys Hooke’s law or that its force is constant, or whatever …)
Hi James — thanks a lot for your comment — there are two very interesting points in it.
You are absolutely right that the force clearly isn’t going to be constant for the whole 0.6m. But, as I assume you know very well given your question, the ‘elastic potential energy method’ that you have used is not going to give the correct answer either. In reality, bows obey a complex force/distance curve which archers call the ‘draw force curve’, which lies somewhere between the ‘Hooke’s Law’ curve you have used and the ‘constant force’ curve that I have used; the energy is the integral of the draw force curve. But in this case the examiner hasn’t told us the shape of the curve! So the best one can do is to state an assumption and go with it. However I think it might well be true that your answer was the expected/desired one, and t he very least my answer above should have included some statement of what I have just said in this comment, to make it clear that there was some reasoning behind it.
Is there error carry forward for PAT? I’m afraid nobody knows the answer to that question. I kind of assume not, because it would be daft, but I guess it is a bit daft to set a question which requires students to integrate a function that you don’t give them, so who knows? I suppose the motto is to be rigorous: show your working clearly, and clearly state any assumptions you have made and why you have made them.
Hope this helps, and the very best of luck tomorrow!
Alright, thank you so much!
For the first part i don’t think it is wise to assume that the force is constant throughout 0.60m. I did the elastic potential energy method and get an alternative solution. Anyway, the answer for the first part will affect the rest of the answers in the last question….. is there error carry forward for PAT? If not does that mean that the whole thing will be wrong if we chose the wrong technique to calculate?
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