Posted by **Brianna** on Friday, May 11, 2012 at 7:54pm.

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?

m/s

(b) If the arrow is shot straight up, how high does it rise?

m

- physics -
**Elena**, Saturday, May 12, 2012 at 4:57am
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.

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