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11 September, 10:18

A 125.0 g arrow is pulled backed 27.0 cm against a bowstring. If the spring constant of the bowstring is 1175 N/m what speed will thw arrow leave the bow

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  1. 11 September, 10:46
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    26.18 m/s

    Explanation:

    From the question,

    The Energy stored in the string = kinetic energy of the arrow.

    1/2 (ke²) = 1/2 (mv²) ... Equation 1

    Where k = spring constant of the bowstring, e = extension, m = mass of the arrow, v = velocity of the arrow.

    make v the subject of the equation

    v = √ (ke²/m) ... Equation 2

    Given: k = 1175 N/m, e = 27 cm = 0.27 m, m = 125 g = 0.125 kg.

    Substitute this values into equation 2

    v = √ (1175*0.27²/0.125)

    v = √ (685.26)

    v = 26.18 m/s
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