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An archer draws his compound bow and shoots an arrow. The 23 g arrow leaves the bow with a velocity of 88 m/s. The power stroke of the bow is 57 cm; that is, the bowstring exerts force on the arrow through a displacement of 57 cm. The peak draw weight of the bow is 312 N. (This is the maximum force that the archer has to exert on the bowstring.)

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Answer:

156.24N

Explanation:

From the data above;

Mass = 23g = 0.023kg

V = 88m/s

Kinetic energy = ½mv²

K.e = ½* 0.023 * 88²

K.E = 89.056J

The force required to accelerate the arrow to that speed and energy in 0.57m

Using equation of motion;

V² = U² + 2as

But U =0

V² = 2as

a = v²/2s

a = 88² / (2*0.57)

a = 6792.98m/s²

The force (F) that accelerates the arrow with that velocity = ?

F = M*a

F = 0.023 * 6792.98

F = 156.24N

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