2. 1© Manhattan Press (H.K.) Ltd. 20.3 Energy stored in stretched wire Elastic potential energy Elastic potential energy per unit volume

3. 2 © Manhattan Press (H.K.) Ltd. 20.3 Energy stored in stretched wire (SB p. 67) Elastic potential energy A wire is stretched, work done is stored as elastic potential energy.

4. 3 © Manhattan Press (H.K.) Ltd. 20.3 Energy stored in stretched wire (SB p. 67) Elastic potential energy (a) Elastic limit is not exceeded

5. 4 © Manhattan Press (H.K.) Ltd. 20.3 Energy stored in stretched wire (SB p. 68) Elastic potential energy (b) wire is stretched beyond elastic limit

6. 5 © Manhattan Press (H.K.) Ltd. 20.3 Energy stored in stretched wire (SB p. 68) Elastic potential energy per unit volume Elastic potential energy per unit volume of wire = ½ x Stress x Strain = Area under stress-strain graph Go to Example 4 Example 4

7. 6 © Manhattan Press (H.K.) Ltd. End

8. 7 © Manhattan Press (H.K.) Ltd. Q: Q: A wire of natural length  and cross-sectional area A is fixed at one end. A mass m is attached to the lower end of the wire and lowered slowly so that it finally hangs in equilibrium. (a) If E is the Young modulus of the wire and the elastic limit is not exceeded, find the extension in the wire in terms of g, , A, m and E. (b) What is the energy stored in the stretched wire? (c) What is the loss in potential energy of the mass? Account for any difference in the answers to (b). (d) Find the maximum extension in the wire if instead of slowly lowering the mass down, it is released when the length of the wire is. Solution 20.3 Energy stored in stretched wire (SB p. 68)

9. 8 © Manhattan Press (H.K.) Ltd. Solution: (a) When the mass hangs in equilibrium, the tension in the wire T = mg. From the definition of Young modulus, (b) Energy stored in the stretched wire (c) Loss in potential energy of mass = Twice the energy stored in the stretched wire Half of the loss in potential energy of the mass is converted into elastic potential energy of the wire. The other half of the loss in potential energy is used to do work against the resistance provided by the hand when the mass is lowered by the hand. 20.3 Energy stored in stretched wire (SB p. 69)

10. 9 © Manhattan Press (H.K.) Ltd. Solution (cont’d): (d) Let e 2 = maximum extension of wire when the mass is released. By the principle of conservation of energy, Loss in PE of mass = Gain in elastic potential energy Note: When the wire is extended by e 2, the mass m is not in equilibrium as T 2 > mg. Subsequently, the mass m will perform vertical oscillation. Return to Text 20.3 Energy stored in stretched wire (SB p. 69)