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

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14