# 17 Joules of Energy From a Mouse Trap

Do you think it would be possible to get 17 joules out of a standard size mouse trap. By my math, it is a torsion coefficient of 3.45 or so out of the spring.

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Clearly 3.45 can't be the value of a torsion coefficient from dimensional analisys alone. – Olin Lathrop Feb 22 '14 at 17:20

I haven't used a mousetrap for several decades, but as I recall the moving arm is about 5cm long, so the tip moves 0.05$\pi$ or about 0.16m. To get 17J of work the force at the tip of the arm would need to be 100N. I'm fairly sure the force isn't anything like that great. I remember being able to pull the arm back with one finger. I would guess the force is nearer 10N, so you'd only get around 2J out.

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Yes, if you burn it. Neglecting the metal, a $25\ g$ wooden mousetrap at $15\ MJ/kg$ should yield about $375\ kJ$.

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I was going to make this a joke comment but then I realized that it's an answer :) – Mark Beadles Feb 15 '12 at 18:21
+1: good answer :-) – John Rennie Feb 15 '12 at 19:54
Nice answer...I should have been a bit more specific. – John Feb 16 '12 at 2:38
And if you throw it into a black hole, you can get about a third of a megaton. I don't think this is in the spirit of the question, but it's cute. – Ron Maimon Feb 16 '12 at 6:08

Good guesstimating John Rennie. This fellow did some measurements (on a rat trap) and got slightly over 3 Joules:

http://www.instructables.com/id/Mouse-Trap-Speed/step4/Analysis-using-Basic-Physics/

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$$U=\frac{1}{2}k\theta^2$$. A standard mouse trap is usually set at $\approx 180^o=\pi\space$ radians. $$\therefore U=0.5*3.45*\pi^2\approx17.025\space J$$ Of course, the torsion law is only an approximation, and it may decrease after use of the moustrap, so the energy obtained would be slightly less.

EDIT: I seem to have misunderstood your question. I don't have any idea what the torsion constant of a spring should be, but 17 joules is pretty small.

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Even though you could not answer my question, thank you for the attempt. – John Feb 15 '12 at 4:53

Took the following data: measured force on 0.045 m trap arm and computed torque

(0 , .1215) (1.57 , .2565) (3.14 , .378)

best fit line to data torque = .0816 * angle + 0.1238

area under this line is energy stored in spring