1. Work, Energy and Power

2. Work and Energy  Work is defined as the transforming or converting from one form of energy into another form of energy.  Every time you flip on a light switch, work is being done. It is changing electrical energy in to heat (light).  Energy is the ability to do work or effect change.

3. Different Forms of Energy  Kinetic energy (motion)  Elastic energy (springs)  Electrical energy (batteries)  Thermal energy (fire)  Radiation energy (light)  Gravitational energy  Chemical energy (food)  Wind energy  Sound energy (sound waves)  Hydraulic energy (waterfalls)  Nuclear energy (atomic nuclei, the sun)

4. Potential Energy vs Kinetic Energy  Potential Energy – is energy that can be stored in an object.  A battery has potential Energy as it has a lot of energy stored in it (electrical) that is just waiting to be used.  Kinetic Energy - is the energy that an object has because of its motion.  When I move I release energy that I had stored up in my body.

5. Potential vs Kinetic Energy  Ex. When you throw a ball up in the air, you give it kinetic energy (it moves).  This energy is transformed into gravitational potential energy as the ball rises.  At the balls highest point, all the kinetic energy has been transformed into potential energy.  As the ball comes back down, gravitational potential energy is transformed into motion which is kinetic energy.

6. Potential vs Kinetic Energy  Kinetic energy has a relationship between mass and velocity. The greater the mass or the speed, the greater the kinetic energy.  KE = 1mv 2 2  Where KE is Kinetic Energy measured in joules, m is mass measured in kg and v is velocity is measured in m/s.

7. Work and Energy  W = KE final – KE initial W = ∆KE  Where W is work and it is measured in joules and KE is kinetic energy measured in joules.

8. Law of Conservation of Energy  Work can then be deduced as a change of energy from one type to another.  Thus W = ∆E  This is known as the Law of Conservation of Energy - Energy can be neither created nor destroyed; it can only be transferred or transformed.  So the amount of energy needed to turn on a light bulb derived from the potential electrical energy passing through it.

9. Power and Watts  Power is the rate at which energy is transformed or, the rate at which work is done. It is measured in Watts.  P = ∆E ∆t Where P is power and is measured in Watts, E is energy measured in Joules and t is time in seconds. 1watt = 1J/s

10. Power and Watts  Since we know that W = ∆E we can substitute W for ∆E  P = W ∆t Where P is power in watts, W is work done in Joules, and t is the time interval in seconds.

11. Kilowatt Hours  Energy is commonly measured in Joules (J), but can also be measured in kilowatt hours (KWh). 1 kWh = 3 600 000 J  KWh is used by B.C. Hydro to calculate your energy bill.

12. Example #1  A stereo has a power rating of 200W & is used for 1800 seconds. How much energy is used? E = Pt E = (200)(1800) E = 360,000 J 360,000 Joules of energy were used to work the stereo

13. Example #2  A video game has a power rating of 175W & is used for 3600 seconds. How much energy is used? E = Pt E = (175)(3600) E = 630,000 J 630,000 Joules of energy were used to work the video game

14. Example #3  A hair dryer has a power rating of 425W & is used for 10 minutes. How much energy is used? E = Pt E = (425)(600) E = 255,000 J 255,000 Joules of energy were used to work the hair dryer

15. Example #4  A stereo plays for 2 hours & 30 min (9000 seconds) & consumes 1, 800, 000 J of energy. What is the amount of wattage used? P = ∆E ∆t P =1,800,000 J 9000s P= 200 J/s or 200 Watts

16. Example #5  A kettle is used for 5 min & has a power rating of 24 Watts. How much work is being done by the kettle? P = W ∆t W = P∆t W = (24 watts)(300s) W = 7,200 J

17. Example #6  A 100 W light bulb is left on for 3 hours. How much energy did it use? Express your answer in scientific notation. 3 h (3600 s) = 10,800 s 1 h P = ∆E ∆E = P(∆t) ∆t∆E = 100 W(10,800 s) ∆E = 1,080,000 ∆E = 1.1 x 10 6

18. Challenging Question  A circuit has a resistance of 5Ω. If the battery powering the circuit has a voltage of 10V, how much power is used to run this circuit?  R=V/I  I= q/t  V=E/q  P=E/t  V=P/I R = 5Ω V = 10 V P= ? I = ? Step 1 R=V/I 5 = 10 / I 5 I = 10 I = 10/5 I = 2 A Step 2 V=P/I 10 = P/2 10 * 2 = P P = 20 watts