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Elastic Energy

Elastic Energy. SPH3U. Hooke’s Law. A mass at the end of a spring will displace the spring to a certain displacement (x). The restoring force acts in springs to try and return the spring to its un-stretched state This force opposes the displacement and follows Hooke’s Law: F = - kx.

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Elastic Energy

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  1. ElasticEnergy SPH3U

  2. Hooke’s Law • A mass at the end of a spring will displace the spring to a certain displacement (x). • The restoring force acts in springs to try and return the spring to its un-stretched state • This force opposes the displacement and follows Hooke’s Law: F = - kx

  3. Hooke’s Law • F = - kx • The – sign can be omitted from problem solving as it only indicates that the force direction is opposite to the displacement direction. • This is not a constant force as force increases with x (so does the acceleration too). • The force of gravity is balanced by the spring force.

  4. Example 1 • A spring stretches 0.150 m when 0.300 kg hangs from it. Find the spring constant. • When the mass hangs from the spring, the weight of the mass is equal to the restoring force. F = |kx| = |mg| k = [(0.300 kg) (9.81 m/s2)]/[0.150 m] = 19.6 N/m

  5. Example 2 • The force constant of a spring is 48.0 N/m. It has a 0.25 kg mass suspended from it. What is the extension of the spring? F = |kx| and F = mg x = 0.051 m • Note: A compressed spring has a “-“displacement.

  6. ElasticPotentialEnergy • Elastic potential energy is stored in an object if there is no net deformation. • This means that the object can return to its original form. • Work put into extending a spring, for example, is equal to the work released by the spring when it returns to its natural shape. (Some or little heat results.) • Other examples are: trampolines, elastic bands, etc.

  7. ElasticPotentialEnergy • F = kx in the elastic region only. • The slope is the spring constant k that varies with material. (N/m)

  8. ElasticPotentialEnergy • The energy stored in a spring is equal to the work done to displace the spring which is represented by the area under the graph for the elastic region. • Area = (1/2) base height = (1/2) F x = (1/2) kxx = (1/2) kx2 • Eel = (1/2) kx2 (again measured in Joules)

  9. Example 1 • Find the elastic potential energy of a spring (k = 160 N/m) compressed 8.0 cm. Eel= (1/2) kx2 = (1/2) (160 N/m) (-0.080 m)2 = 0.51 J

  10. Example 2 • A 0.50 kg hockey puck slides at 15.0 m/s. It hits a spring loaded bumper (k = 360 N/m); how much does the bumper spring compress at maximum compression? • The puck loses Ek to do work to compress the spring; - ∆Ek = ∆Eel -(0 - (1/2) mv2) = (1/2) kx2 - 0 x = ± 0.56 mAs compressed, x = -0.56 m

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