Aspring with spring constant 450 n/m is stretched by 12 cm. what distance is required to double the amount of potential energy stored within the spring?

The distance required = 16.97 cm

Explanation:

Hook's Law

From Hook's law, the potential energy stored in a stretched spring

E = 1/2ke² Equation 1

making e the subject of the equation,

e = √(2E/k) Equation 2

Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.

Given: k = 450 N/m, e = 12 cm = 0.12 m.

E = 1/2(450)(0.12)²

E = 225(0.12)²

E = 3.24 J.

When the potential energy is doubled,

I.e E = 2×3.24

E = 6.48 J.

Substituting into equation 2,

e = √(2×6.48/450)

e = √0.0288

e = 0.1697 m

e = 16.97 cm

Thus the distance required = 16.97 cm

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