Posted by Brianna on .
An archer puts a 0.26-kg arrow to the bow-string. An average force of 200 N is exerted to draw the string back 1.4 m.
(a) Assuming no frictional loss, with what speed does the arrow leave the bow?
(b) If the arrow is shot straight up, how high does it rise?
W =F•x =200•1.4 =280 J,
W =KE = m•v²/2,
v =sqrt(2•W/m) = sqrt(2•280/0.26) =46.4 m/s.
KE =PE = m•g•h
h = KE/m•g = 280/0.26•9.8 ≈110 m.
Sometimes I’ve found the answers of this problem based on the following solution
PE =kx²/2 = KE =mv²/2.
This solution gives v =sqrt(k•x²/m) = sqrt(F•x²/x•m) = sqrt(F•x/m) = 32.8 m/s.
And the height from m•g•h =m•v²/2 is
h =v²/2•g = 54.95 m.
But I believe that this solution is incorrect because we have given the magnitude of the average (!) force
So the first solution is quite the thing.