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WORK AND ENERGY. The work done by a force is the product of the magnitude of the force and the distance moved in the direction of the force Recall the.

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Presentation on theme: "WORK AND ENERGY. The work done by a force is the product of the magnitude of the force and the distance moved in the direction of the force Recall the."— Presentation transcript:

1 WORK AND ENERGY

2 The work done by a force is the product of the magnitude of the force and the distance moved in the direction of the force Recall the following basic definitions: i.e. Work Done = F s Potential Energy, P.E. = m g h Kinetic Energy, K.E. = 1212 m v 2m v 2

3 Example 1: A ball of mass 0.6 kg is dropped from a height of 4m above the ground. Find the speed at which the ball hits the ground. 4m Zero P.E. level A B Using the principle of conservation of Mechanical Energy ( P.E. + K.E. ) at A = ( P.E. + K.E. ) at B (0.6) (9.8) (4) + 1212 (0.6) v 2 0 = 0 + 78.4 = v 2 Speed = 8.85 ms –1. ( 3 sig.figs.) v = 8.85 P.E. = m g h K.E. = 1212 m v 2m v 2 Using:

4 Example 2: A particle of mass 0.5 kg is projected up the line of greatest slope on a smooth plane AB inclined at 30° to the horizontal. The particle is given an initial speed 14 ms –1 up the slope from A, and comes to rest at B. Find the distance AB. Using the principle of conservation of Mechanical Energy: ( P.E. + K.E. ) at A = ( P.E. + K.E. ) at B Zero P.E. level 0 + (0.5) 14 2 = 1212 (0.5) (9.8) (d Sin 30°) +0 If the particle moves a distance d: 30° d h h =d Sin 30° 2.45 d d = 20 i.e. Distance AB = 20m B 30° A 49 =

5 Example 3: Two particles A and B, of masses 3kg and 7kg respectively, are connected by a light inextensible string, passing over a smooth fixed pulley. The system is held with the string taut and released from rest, with particle B, 0.5m above the ground. Find the speed in terms of g at which B hits the ground. A B 7g 3g 0.5m Zero P.E. level for B Zero P.E. level for A Total initial ( P.E. + K.E. ) = Total final ( P.E. + K.E. ) 0 + 0 + 0 + 0 = 3g (0.5) + 1212 3v 2 1212 7v 2 –7g (0.5) + Speed, v = ms –1. Using the principle of conservation of Mechanical Energy: 0 = 5v 2 – 2g

6 Example 4: A particle of mass 2 kg slides down a rough slope AB of length 4m inclined at 30º to the horizontal, against a resistive force of N. The particle starts from rest at A, find its speed at B. g2g2 N a 2g 30° g2g2 ( P.E. + K.E. ) at A + Work Done = ( P.E. + K.E. ) at B 30° 4 h 4 Sin 30° h = = 2 30° A B Zero P.E. level 2 g (2) + 0 + g2g2 ( – 4) = 0 + 1212 (2) v 2 Speed, v = ms –1. 4g – 2g = v 2

7 Summary of key points: This PowerPoint produced by R.Collins ; Updated Aug. 2009 Work Done = F s P.E. = m g h K.E. = 1212 m v 2m v 2 The principle of conservation of Mechanical Energy: Initial ( P.E. + K.E. ) + Work Done = Final ( P.E. + K.E. )


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