WORK, ENERGY AND POWER

1. WORK, ENERGY AND POWER WHY ARE WE WORKING SO HARD?

2. GET the FACTS. • Work is done when a force acts on an object and it moves in the direction of the force. work done W = F.Δx.cosӨ • Work is a scalar quantity. • The unit for work is the joule (J) or the newton metre (N.m). • 1 joule = 1 N.m = 1 kg.m.s-2. m = 1 kg.m2.s-2

3. If a force acts on an object but the object does not move, no work is done on the object. • If an object moves at constant velocity across a frictionless surface, no force acts on the object and so no work is done.

4. Force Weight This man is doing work against the force of gravity, but is doing no horizontal work on the suitcase, because he is exerting a force vertically and not horizontally.

5. WORK DONE BY FORCES ACTING AT AN ANGLE TO THE HORIZONTAL The force doing the work and the motion must be in the same direction. If the force doing the work acts at an angle, first determine the horizontal component of the force acting at an angle. Use the horizontal component of the force to determine the amount of work done.

6. Fapplied Fvertical θ θ Fhorizonal • FH = F cos θ, where θ is the angle between the applied force and the horizontal. • work done = Fhorizontal .Δx where Δx = distance moved

7. The Inuit man pulling the sledge exerts a force of 200N and moves the sledge 20m along the level ground. The rope is inclined at an angle of 30o to the horizontal. • Determine the work done by the man, while he moves the penguins.

8. The horizontal component of the force can be calculated as follows: • FH = F cos 30o = 173,21N • Work done = F.Δx • = 173,2 X 20 • = 3 464,10J • OR W = FΔx.cosӨ • = 200(20).cos 30° • = 3464,10J

9. Work done on an inclined plane • The diagram is important and learners must be able to draw:

10. ENERGY Energy is defined as the capacity to do work. The energy of an object or a system is determined by calculating the amount of work that it can do. The unit of energy is the joule (J) and it is a scalar quantity. Energy is transferred when work is done.

11. The object doing the work loses energy. • The object on which work is done, gains energy. • When an object is lifted in the earth’s gravitational field, the work is done against the force of gravity. • When an object is moved across a rough surface, the work is done against the force of friction.

12. THE LAW OF CONSERVATION OF MECHANICAL ENERGY Mechanical energy (or any other energy) cannot be created or destroyed, only changed from one form to another. The mechanical energy of an object is the sum of it’s gravitational potential energy (Ep) and it’s kinetic energy (Ek).

13. Gravitational Potential Energy Gravitational Potential energy is the energy an object possesses due to its position in the earth’s gravitational field. Ep = mgh m = mass g = gravitational acceleration h = height above the earth An object gains gravitational potential energy when it is lifted up in the earth’s gravitational field.

14. Deriving an equation for calculating gravitational potential energy Lifting an object at constant velocity to a height h: • Force applied = weight of the object = mg • Distance (s) = h • Work = F.s = gain in Ep • Ep = mgh h Fg = mg

15. Deriving an equation for calculating Ek • A force (F) accelerates an object across a frictionless surface for a distance s. F s • Work done = F.s • Fres = m.a

16. Work done = m.a.s (m x a x s) The object accelerates, velocity increases and the energy is transferred as Ek. • For this movement: • vf2 = vi2 + 2as and vi = 0 • as = ½ vf2 • Ek = ½ mvf2

17. Conservation of energy during free fall Mechanical energy = Ep = mgh X • Mechanical energy = Ep + Ek = mg(h – x) + ½ mvf2 • = Ep at top h Mechanical energy = Ek = ½ mv2 = Ep at top

18. Conservation of energy during the swing of a pendulum

19. At the top of its swing the bob stops for an instant, and so has no kinetic energy. The total energy at the top of the swing is given by Ep = mgh (where h is the height above the lowest point) At the bottom of the swing Ep = 0 and all the energy is kinetic energy. Ek = ½ mv2. v is the maximum speed of the bob.

20. Ep at the top = Ek at the bottom mgh = ½ mv2 v2 = 2gh v = √2gh v2 = 2gh v = √2gh V is the maximum velocity of the bob and occurs at the bottom of its swing.

21. POWER • Power is the rate at which work is done. • p = work done/t or energy transferred/t • The unit for power is joule per second, or J/s, also called a watt (W)

22. Apply your knowledge A boy of mass 40kg is traveling at 8m.s-1 on a skateboard when he reaches a ramp of height 1,2m and of length 5m. A constant force of friction of 16N acts between the skateboard and the surface of the ramp. Calculate his velocity when he reaches the top of the ramp.

23. This is how your answer should look: Ek at the bottom of the ramp = ½ mv2 Ek = ½ x 40 x (8)2 = 1280J This is the total amount of energy at the beginning Ep gained during journey up ramp = mgh Ep gain = 40 x 10 x 1,2 = 480J Energy lost due to friction = F.s = 16 x 5 = 80J Ek remaining = 1280 – (480 + 80) = 720J 720 = ½ mv2 720 = ½ x 40 x v2v = 6 m.s-1

24. Apply your knowledge 2 A woman, shopping in a hurry, applies a force of 60N and moves her trolley at a constant speed of 3m.s-1. What is her power?

25. This is how your answer should look: Work done = Fcosθ Because the force is horizontal, cos θ = 1 Work done = f.s and P = work done/t power = f.s/t but s/t = v power = f.v = 60 x 3 = 180W DO NOT CONFUSE POWER (P) AND MOMENTUM (p). RATHER WRITE OUT “POWER. “