Answer: Approximately 22 cm
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Explanation:
The unstretched spring is 12.0 cm long. When adding a load of 5.0 N, it stretches to 15.0 cm. This is a displacement of 15.0 - 12.0 = 3.0 cm, which is the amount the spring is stretched.
Convert this displacement to meters (so that it fits with the meters unit buried in Newtons).
3.0 cm = (3.0)/100 = 0.03 m
Apply Hooke's Law to find the spring constant k
F = -kx
5.0 = -k*(0.03)
k = -(5.0)/(0.03)
k = -166.667 approximately
Now we must find the displacement x when F = 15 newtons
F = -kx
-kx = F
x = F/(-k)
x = -F/k
x = -15/(-166.667)
x = 0.089 approximately
x = 0.1
The displacement to one decimal place is about 0.1 meters, which converts to 100*0.1 = 10 cm
So the spring will be stretched to about 12cm+10cm = 22 cm
