physics
posted by Brianna on .
An archer puts a 0.26kg arrow to the bowstring. 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?
m/s
(b) If the arrow is shot straight up, how high does it rise?
m

1.
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.
2.
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.