Question

A spring 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?

Answer 1

Answer:

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

Answer 2

The distance required to double the amount of potential energy stored within the spring is mathematically given as

e = 0.1697 m

What distance is required to double the amount of potential energy stored within the spring?

Question Parameter(s):

A spring with spring constant 450 N/m is stretched by 12 cm.

Generally, the equation for Hook's Law   is mathematically given as

E = 1/2ke^

Therefore

e = √(2E/k).

E = 1/2(450)(0.12)^2

E = 3.24 J.

For e potential energy

E = 2*3.24

E = 6.48 J

In conclusion, Subbing

$e = \sqrt{(2*6.48/450)}$

e = √0.0288

e = 0.1697 m

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