Practice

½kx² Formula – Spring / Elastic Potential Energy (Practice)

U = ½kx² — Elastic potential energy of an ideal spring

An ideal spring stores elastic potential energy U = ½ k x², where x is stretch or compression from equilibrium. Doubling x multiplies U by four.

Guide — how to use this formula

1. 1. Measure x from the unstretched / equilibrium length.
2. 2. Convert x to metres; k in N/m.
3. 3. Compute U = ½ k x², or rearrange for x or k.
4. 4. For launches, set U = KE when frictionless.

Derivation

Hooke’s law says the restoring force grows linearly with stretch. Work to stretch from 0 to x is the area under the F–x triangle:

1. Force at stretch x:
   F = kx
2. Average force from 0 to x is ½kx, so work is
   W = \tfrac{1}{2}(kx)\, x = \tfrac{1}{2}kx^{2}
3. That work is stored as elastic PE:
   U = \tfrac{1}{2}kx^{2}

Don’t confuse F = kx (force) with U = ½kx² (energy). Using Fx overcounts by a factor of two.

Examples

Example: k = 200 N/m, x = 0.10 m

U = ½(200)(0.10)² = 1.0 J.

Example: find x from U

k = 100 N/m, U = 2 J → x = √(2U/k) = √0.04 = 0.20 m.

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