Answer:
v=39.05 m/s
Explanation:
Given that
x= 56 cm
F= 158 N
m= 58 g = 0.058 kg
Lets take spring constant = k
At the initial position,before releasing the arrow
F= k x
By putting the values
F= k x
158= 0.56 k
k=282.14 N/m
Now from energy conservation
Lets take final speed of the arrow after releasing
[tex]\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2[/tex]
k x²=mv²
282.14 x 0.56² = 0.058 v²
v=39.05 m/s