Question

11 An unstretched spring is 12,0 cm long. A load of 5.0N stretches it to 15.0cm. How long will it be under a load

Answer 1

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