If we treat a bow as a spring, then by drawing the arrow back against the bow st

Question

If we treat a bow as a spring, then by drawing the arrow back against the bow string, potential energy is stored in the bow material. When released, the potential energy of the bow is transformed into kinetic energy of the arrow. Thus we can use conservation of energy to estimate the velocity of the arrow upon release. a. Let's say that an archer draws back the arrow 0.48 m and requires 150 N of force to maintain the arrow in this position. Upon release of the arrow, what will be the speed of the arrow if it has a mass of 0.02 kg. List any assumptions you must make for this simple analysis.

Explanation / Answer

The first step is to find the composition of the initial and final energies. In this case, the entire problem takes place horizontally and therefore the gravitational potential energy is constant at all times and can be ignored. The initial energy (when the bow is drawn all the way back) is elastic potential energy, and the kinetic energy is zero as the system is at rest. E = U.

On the other hand, once released, the bow converts this initial energy into kinetic energy as the string nears its equilibrium position. At the point of equilibrium for the string, the energy is completely transferred to kinetic energy. E = K.

Ei = U + K (v = 0)

Ei = U + 0

Ei = 1/2 kx2

Ef = U + K (x = 0)

Ef = 0 + K

Ef = 1/2 mv2

Ei = Ef

1/2 kx2 = 1/2 mv2

k = F/x (Hooke’s Law)

1/2 (F/x) x2 = 1/2 mv2

v = (Fx/m)1/2

v = (150*0.48/0.02)1/2

v = 60 m/s

Speed of arrow is 60 m/s

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