Question

An arrow, starting from rest, leaves the bow with a speed of 14.5 m/s. If the average force exerted on the arrow by the bow were doubled, all else remaining the same, with what speed would the arrow leave the bow?

Answer 1

\[W = Fd\] so if the just force is doubled the work done is doubled, \[W = \Delta KE\] so if the work is doubled, the KE is doubled \[KE = (1/2)mv ^{2}\] solve for v with KE and solve for v with 2KE, then compare v's
\[v_{1} = \sqrt{2KEm}, v _{2} =\sqrt{4KEm}\]
so the ratio of v's would be:
\[v _{2} = (2/\sqrt{2})v _{1}, or, v _{2} = (2\sqrt{2}/2)v _{1}, or, v _{2} = 1.41v _{1}\]

Answer 2

Since I failed to notice that the problem provided an actual velocity of the bow, the new velocity would be 20.51 m/s.