Question on kinetic energy transferred by a bow

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The discussion centers on calculating the speed of arrows released from a bow using a force of 1000 N. To determine the speed of a 1 kg and a 2 kg arrow, one must consider the spring constant of the bow and the distance it is drawn. The potential energy (PE) stored in the bow can be calculated using the formula PE = (1/2)kx², where k is the spring constant and x is the distance drawn. This potential energy can then be equated to the kinetic energy of the arrows to solve for their respective speeds.

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i tense a bow holding the string with a dinamometer till it shows a force of 100 kg or 1000 N to be more correct

at what speed would be released a 1 kg arrow and a 2 kg arrow?
 
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I would think you'd need some kind of spring constant for the bow (assuming it is constant over it's entire range of travel) and the distance the bow is exteded so you could calculate the potential energy in the bow in the extended position. You are given a hint at the PE by being given the force required to pull the bow, but that gives you F= kx, not kx² which you need for the PE. At that point you could equate it to kinetic energy transferred to the arrow and then solve for the speeds.
 

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