Question on kinetic energy transferred by a bow

[eosphorus]

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?

 

[FredGarvin]

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.

 

[quicknote]

The kinetic energy transferred by a bow is dependent on the speed and mass of the arrows being released. Using the equation KE = 1/2mv^2, we can calculate the kinetic energy for each arrow.

For the 1 kg arrow, the kinetic energy would be equal to 500 J if released at a speed of 10 m/s. For the 2 kg arrow, the kinetic energy would be equal to 1000 J if released at the same speed of 10 m/s.

However, it is important to note that the force of the bow and the speed of the arrows may not be directly proportional. Factors such as the elasticity of the bow and the friction between the bowstring and the arrow can affect the speed of the arrow. Therefore, it is difficult to determine the exact speed at which the arrows would be released without conducting an experiment.