1. Elastic Energy 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 The “k” is a constant for each material in Nm -1.

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 when a mass is at rest hanging from the spring.

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 ms -2 )]/[0.150 m] = 19.6 Nm -1

5. Example 2 The force constant of a spring is 48.0 Nm -1. 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. Elastic Potential Energy 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. Elastic Potential Energy F = kx in the elastic region only. The slope is the spring constant k that varies with material. (Nm -1 )

8. Elastic Potential Energy 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) kx 2 E el = (1/2) kx 2 (again measured in Joules)

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

10. Example 2 A 0.50 kg hockey puck slides at 15.0 ms -1. It hits a spring loaded bumper (k = 360 Nm -1 ); how much does the bumper spring compress at maximum compression? The puck loses Ek to do work to compress the spring; - ∆E k = ∆E el -(0 - (1/2) mv 2 ) = (1/2) kx 2 - 0 x = ± 0.56 m As compressed, x = -0.56 m