1. Energy and work Sections 12, 13, 14 and 15 http://phet.colorado.edu/en/simulation/energy-skate-parkhttp://phet.colorado.edu/en/simulation/energy-skate-park Interactive website

2. Aims Be able to use the equation Kinetic energy = ½ mV 2 Be able to use the equation  PE = mg  H Investigate and apply the conservation of energy Be able to use the equation  W = F  S, including force not along the line of motion

3. Section 12 Energy symbol E unit Joules Energy is the ability to do work. Energy can take many forms: heat, electrical, kinetic, potential etc… Whenever a mass moves it has kinetic energy: Example: Calculate the kinetic energy of a car of mass 900 Kg travelling at 20 ms -1 E k =0.5mV 2 E k = 0.5 x 900 x (20) 2 E k = 180 000 Joules The equation for kinetic energy is:

4. Section 13 Change in gravitational potential energy Example: A man of mass 70 kg climbs a ladder and moves through a vertical height of 4.5 m. Calculate the change in gravitational potential energy.  Egrav = mg  h  Egrav = 70 x 10 x 4.5  Egrav = 3150 J If a mass is lifted vertically in a gravitational field it gains gravitational potential energy

5. Section 15 Work symbol W unit Joules Work is the amount of energy converted from one form into another or the amount of energy used (e.g. kinetic in potential) Whenever energy changes from one form into another work is done. Mechanical work is given by: Distance moved Force applied Example: A force of 20 N moves an object a distance of 300 m. Calculate the work done. Work = force x distance Work = 20 x 300 Work = 6000 Joules What happens to the energy (6000 J) if there is no friction? All the energy is converted into kinetic energy. Hence K.E. = 6000 J What happens to the energy (6000 J) if there is friction? Only some of the energy is kinetic energy the rest is converted into heat, sound etc.

6. The force and the distance have to be parallel for the equation to work. If they are not then you need to find the component of the force parallel to the direction of movement. Force applied Force F Force F can be replaced by two components  One vertically: F x Sin  The other horizontally: F x Cos  Direction of movement

7. F x Cos  Distance moved d Work done by force F is: Work = F x Cos  x d Example: If F = 200 N (  = 40 o ) moves the object horizontally a distance of 50 m, calculate the work done by the force F Force F  Work = F x Cos  x d Hence Work = 200 x Cos 40 x 50 Work = 7660 Joules

8. Section 14 Conservation of energy Energy can neither be created nor destroyed. It can only be converted from one form into another. For most systems ignoring friction: A B C If the yellow object moves from A to B to C the total energy is fixed. Throughout the journey the sum of kinetic energy and potential energy is a constant

9. Work is the amount of energy converted from one form to another. If friction acts then the work done by friction can be calculated using: force x distance The previous equation is now altered to take the frictional force into account and is rewritten as : Example A stone is thrown upwards with a kinetic energy of 200 J, it rises to a height (24 m) such that it gains 140 J of gravitational potential energy. How much work is done and what is the size of the frictional force. Using the above equation: KE b + PE b = KE a + PE a + work done 200 + 0 = 0 + 140 + Work done Hence Work done = 60 Joules Work done = Force x distance The distance is 24 m and the work done is 60 J Hence Force = 60/24 = 2.5 Newton

10. Be able to use Kinetic energy = ½ mV 2 Calculate the kinetic energy of a car (1200 kg) moving with a speed of 20 m s -1. Answer: 240 000 Joules

11. Be able to use the equation  PE = mg  H Elevations are in metres and g = 9.81 m s -2 A ball of mass 0.10 Kg rolls down the hill. Calculate the change in gravitational potential energy. Answer: -12.753 Joules

12. Be able to use the equation  W = F  S (example 1) The tugboat pulls on the ship with a force of 30 000 N, it maintains this force over a distance of 400 metres. Calculate the work done. Answer: 12 000 000 Joules

13. Two tug boats each providing a force of 20 000 N are pulling a ship. The angle between the direction of movement and the ropes is 20 o in each case. Calculate the work done by each tugboat if the ship is pulled a distance of 1200 metres. Answer: 22552623 Joules. Be able to use the equation  W = F  S (example 2)

14. Apply the conservation of energy A man of mass 80 Kg is placed in the cradle and given a velocity of 15 m s -1 upwards. If there is no friction what height will he reach? Answer: 11.46 m If he reaches a height of 9.5 metres, what was the size of the frictional force? Answer: (-) 162 Newton