An archer pulls back 0.75 m on a bow which has a stiffness of 200 N/m. The arrow weighs 50 g. What is the velocity of the arrow immediately after it crosses the bowstring (traveled 0.75 m)?
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The Potential Energy of the bow is converted to the Kinetic Energy if the arrow, so:
E P = E K
2 1 k x 2 = 2 1 m v 2
v = x m k = ( 0 . 7 5 ) ( 0 . 0 5 2 0 0 ) ≈ 4 7 . 4 s m