If we treat a bow as a spring, we can use conservation of energy concerns to est

Question

If we treat a bow as a spring, we can use conservation of energy concerns to estimate the velocity of the arrow. By drawing the arrow back against the bow string, potential energy is stored in the bow material. Upon release of the arrow, that potential energy is transformed into kinetic energy of the arrow. Let's say that an archer draws back the arrow 0.48 m and requires 160 N of force to maintain the arrow in this position. When the arrow leaves the string, what will be the speed of the arrow if its mass is 0.025 kg. List any assumptions you must make for this simple analysis.

Explanation / Answer

For spring energy

F= Kx

160 = K*0.48

k =160/0.48 = 333.3 N/m

By energy conservation, spring energy get converted into kinetic energy

0.5 k x^2 = 0.5mv^2

333.3*0.48*0.48 = 0.025*v^2

v^2 = 76.8/0.025 = 3072

v = 55.4 m/s

Assumption is final KE of bow is zero. All energy is harnessed in arrow

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