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Elasticity

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  1. Elasticity A property of matter that enables an object to return to its original size and shape when the force that was acting on it is removed.

  2. What can you observe in this figure?

  3. Hooke’s Law State that: “The extension of a spring is directly proportional to the stretching force acting on it provided the elastic limit of the spring is not exceeded”.

  4. Elastic limit: The maximum stretching force which can be applied to the spring before it ceases to be elastic.

  5. x F = mg The mathematical expression for Hooke’s law : F = k x F = Force on the spring x = extension k = Force constant of the spring (Nm-1) Force constant of a spring:the force that is required to produce one unit of extension of the spring.

  6. 20 cm 26 cm x F = mg Example 1 A spring has an original length of 20 cm. with a load of mass 300 g attached to it, the length of the spring is extended to 26 cm. • Calculate the spring constant. • What is the length of the spring when the load is • increased by 200 g? (assume that g=10 Nkg-1).

  7. Solution lo = 20 cm l1 = 26 cm x = l1 – lo = 26 – 20 = 6 cm = 0.06 m m = 300 g = 0.3 kg g = 10 N kg-1 F = mg = 0.3 x 10 = 3 N k = F/x = 3 / 0.06 = 50 Nm-1 1 m = 300 g + 200 g = 500 g = 0.5 kg F = mg = 0.5 x 10 = 5 N x = F / k = 5 / 50 = 0.1 m = 10 cm l1 = lo + x = 20 cm + 10 cm = 30 cm Length of the spring is being 30 cm 2

  8. Elastic Potential Energy The elastic Potential Energy is the energy stored in a spring when it is extended or compressed. The result of the work done to extend or compress the spring.

  9. KATAPULT ARROW

  10. Ep = Elastic Potential Energy (J) k = Force Constant of the spring (N/m) x = extension of the spring (m)

  11. x F Mass of the ball (m) = 300 g The spring constant (k) = 200 Nm-1. Spring extension (x) = 5 cm What is the maximum velocity of the ball when the stretching force is released? Example 2

  12. Solution Maximum kinetic energyis equal toelastic potential energy ½ m v2= ½ k x2 m = 300 g = 0.3 kg x = 5 cm = 0.05 m V2 = (k x2) / m V2 = (200 Nm-1 x (0.05 m)2 ) / 0.3 kg V2 = ( 200 x 0.0025) / 0.3 V2 = (0.5) / 0.3 = 1.666 V = 1.29 ms-1

  13. Factors that effect the elasticity

  14. APPLICATION IN DAILY LIFE

  15. Summary

  16. In this lesson, we learnt that: • Forces can change the shape of an object. • Objects that return to their original shape when the forces acting on them are removed are elastic. • Hooke’s Law state that the force on a spring is directly proportional to its extension, that is: F = k x F= force (N) K = spring constant (Nm-1) x = spring extension (m)

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