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

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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hope this !