Elastic Potential Energy

When a spring or elastic string is stretched or compressed, it stores elastic potential energy (EPE). This energy can be converted to kinetic energy and vice versa, providing a rich source of mechanics problems.

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1 Hooke's Law

T = (lambda / l) * x = kx

where:

• T is the tension in the spring/string
• lambda is the modulus of elasticity (a property of the spring)
• l is the natural length
• x is the extension (or compression)
• k = lambda/l is the stiffness (spring constant)

Hooke's Law holds for extensions up to the elastic limit.

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2 Elastic Potential Energy Formula

The work done in stretching a spring from natural length by extension x is:

EPE = (lambda x^2) / (2l) = (1/2) k x^2

This is the area under the force-extension graph (which is a straight line through the origin for a Hooke's Law spring).

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3 Derivation

The tension at extension s is T = (lambda/l) s.

Work done = integral from 0 to x of T ds = integral from 0 to x of (lambda/l) s ds = (lambda/l) [s^2/2] from 0 to x = lambda x^2 / (2l)

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4 Worked Example 1

A spring has natural length 0.5 m and modulus of elasticity 40 N. Find the EPE stored when it is stretched by 0.2 m.

Solution:

EPE = lambda x^2 / (2l) = 40(0.2)^2 / (2 * 0.5) = 40(0.04) / 1 = 1.6 J

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5 Worked Example 2

A spring with k = 200 N/m is compressed by 0.1 m. Find the EPE.

EPE = (1/2)(200)(0.1)^2 = (1/2)(200)(0.01) = 1 J

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6 Conservation of Energy with Springs

When a particle attached to a spring moves, energy is conserved (in the absence of friction):

KE + GPE + EPE = constant

Worked Example 3

A 0.5 kg mass is attached to a spring (k = 50 N/m, natural length 0.3 m) on a smooth horizontal surface. It is pulled 0.2 m beyond the natural length and released from rest. Find the speed when it passes through the natural length position.

Solution:

At maximum extension: KE = 0, EPE = (1/2)(50)(0.2)^2 = 1 J.

At natural length: EPE = 0, KE = 1 J.

(1/2)(0.5)v^2 = 1, v^2 = 4, v = 2 m s^(-1)

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7 Worked Example 4 -- Vertical Spring

A 2 kg mass hangs from a spring (lambda = 100 N, l = 0.5 m). It is pulled down 0.1 m from equilibrium and released. Find the speed at equilibrium.