Question on kinetic energy transferred by a bow

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To determine the speed at which a 1 kg and a 2 kg arrow would be released from a bow exerting a force of 1000 N, one must consider the potential energy stored in the bow. This requires knowing the spring constant of the bow and the distance it is drawn, as the potential energy is calculated using the formula PE = 1/2 kx². The force given allows for the calculation of the spring constant using F = kx, but further information about the draw distance is necessary. Once the potential energy is established, it can be equated to the kinetic energy of the arrows to solve for their release speeds. Accurate calculations hinge on these parameters to ensure precise results.
eosphorus
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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.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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