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Understanding Work, Energy, Power & Efficiency

Understanding Work, Energy, Power & Efficiency. Energy. Work done is equal to the amount of energy transferred. Forms of energy:. Kinetic energy Gravitational potential energy Elastic potential energy Sound energy Heat energy Light energy Electrical energy Chemical energy.

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Understanding Work, Energy, Power & Efficiency

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  1. Understanding Work, Energy, Power & Efficiency

  2. Energy

  3. Work done is equal to the amount of energy transferred. Forms of energy: • Kinetic energy • Gravitational potential energy • Elastic potential energy • Sound energy • Heat energy • Light energy • Electrical energy • Chemical energy

  4. Energy – Work Done When energy changes from one form to another, Work is done. Energy transfer when the work is done.

  5. Examples A student pushes a trolley of books. Chemical energy stored in the body of the student is transferred to the trolley as kinetic energy. The farmer lifting a bag of rice onto the lorry transfer energy to the bag as gravitational potential energy.

  6. Only moving objects possess kinetic energy.

  7. The kinetic energy of the object is the same as the work done to change the velocity of the object from 0 m s-1 to v m s-1.

  8. Initial velocity, u = 0 m s-1 Final velocity, v = v m s-1 Displacement, s = s m Work done, W = F x s = m a s W = ma (v2/2a) W = ½ m v2 From equation of linear motion: V2 = u2 + 2as S = (v2) / 2a The kinetic energy, Ek of an object of mass, m kg traveling at a velocity of v m s-1 is given as: Ek = ½ m v2

  9. EXAMPLE 1 A car of mass 950 kg accelerates from a velocity of 20 m s-1 to a velocity of 35 m s-1. What is the work done for the car to accelerate? v = 20 ms-1 v = 35 ms-1 Solution: m = 950 kg, u = 20 ms-1, v = 35 ms-1 Work done, W = Change in kinetic energy of the car W = Final Ek – Initial Ek W = ½ mv2 – ½ mu2 = ½ m (v2-u2) W = ½ x 950 ms-1 x (35 ms-1 – 20 ms-1) = 391875 J

  10. EXAMPLE 2 What is the kinetic energy of a man of mass 65 kg jogging at a velocity of 3 m s-1 Solution Mass, m = 65 kg, velocity, v = 3 m s-1 Kinetic energy, Ek = ½ mv2 Ek = ½ x 65 kg x (3 m s-1)2 Ek = 292.5 J

  11. box h When an object of mass, m kg is raised to a height, h meters above the earth’s surface, the object possesses gravitational potential. The gravitational potential energy is equal to the work done to raise an object to a particular height. Work done, W = F x s W = mgh Gravitational potential energy, Ep = W Ep = mgh

  12. box B 50 N 4 m F= 25 N h = 2 m A 30o EXAMPLE 1 Figure below shows a box of weight 50 N being pulled up on a smooth inclined plank by a force of 25 N from A to B. • Calculate the work done by the 25 N force. • What is the work done against gravity? • State the value of the gravitational potential energy of the box at B.

  13. m = 2 kg h = 1.2 m EXAMPLE 2 Calculate the gravitational potential energy in respect of each of the following. Solution Potential energy of weight, Ep = mgh Ep = 2 kg x 9.8 ms-2 x 1.2 m Ep = 23.52 J

  14. The total amount of energy always remains the same.

  15. EXAMPLE 1 In a softball game, a ball was miss hit and flew vertically upwards with an initial velocity of 15 ms-1. what is the maximum height attained by the ball? Solution If air resistance is neglected, we can assume that the kinetic energy of the ball is changed to gravitational potential energy only. ½ m v2 = mgh ½ x m x (15 ms-1) 2 = m x 9.8 ms-2 x h h = (15 ms-1)2 / (2 x 9.8 ms-2) h = 11.48 m

  16. EXAMPLE 2 • A durian of 2.5 kg falls from a height of 12 m to the ground. • What is the loss of potential energy when the durian is 4 meters above the ground? • Calculate the velocity of the durian at this height. Solution • Loss of potential energy = m g (12 m) – m g (4 m) • = m g (12 m – 4 m) = 2.5 kg x 9.8 ms-2 x 8 m • = 196 J • Gain in kinetic energy = Loss of potential energy • = 196 J • ½ x 2.5 kg x v2 = 196 J • v2 = (2 x 196 J) / (2.5 kg) = 156.8 • v = 12.5

  17. Quiz

  18. u = 2 m s-1 v = 6 m s-1 3 m • Calvin rides his bicycle down the slope of a hill 3 m high at an initial velocity of 2 m s-1, without pedaling. At the foot of the hill, the velocity is 6 m s-1. given that the mass of Calvin with his bicycle is 75 kg, find • The initial kinetic energy of the bicycle, • The initial potential energy of the bicycle, • The work done against friction along the slope.

  19. Solution a. Initial kinetic energy, Ek = ½ m u2 Ek = ½ (75 kg) x (2 m s-2) = 150 J b. Initial potential energy, Ep = mgh Ep = (75 kg) x (9.8 m s-2) x (3 m) Ep = 2205 J c. Kinetic energy at the foot of the hill, Ek = ½ m v2 Ek = ½ (75 kg) x (6 m s-1)2 Ek = 1350 J Work done against friction along the slope, W = total initial energy – kinetic energy at the foot of the hill W = (150 + 2205) – 1350 W = 2355 – 1350 = 1005 J

  20. Summary

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