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An archer puts a 0.26 kg arrow to the bowstring. An average force of 184.5 N is exerted
to draw the string back 1.26 m.
The acceleration of gravity is 9.8 m/s
Assuming no frictional loss, with what
speed does the arrow leave the bow?
Answer in units of m/s

  • physics -

    (Average force) x (draw back distance) = work done on bow

    Set that equal to the arrow's kinetic energy in flight. Solve for V

    You will not need to use the acceleration of gravity.

    (M/2)*V^2 = 184.5*1.26 = 232.5 J
    V = sqrt(2*232.5/0.26)= 42.3 m/s

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