1. 1 Work, Energy & Power Chapter 10

2. 2 Lesson 1 Feb 10 Specific Instructional Objectives At the end of the lesson, students should be able to: –Show understanding of the Physics concept of Work –Correctly identify Work from given situations –Recall and show understanding of the formula to calculate work done –Solve related problems involving work –DO NOW: –If you push on a 5.00 kg block with 20.0 N of force, how fast (frictionless) will it be moving after being pushed 2.00 meters?

3. 3 Physics concept of WORK WORK is done only when a constant force applied on an object, causes the object to move in the same direction as the force applied.

4. 4 Physics concept of WORK What IS considered as work done in Physics: –You push a heavy shopping trolley for 10 m –You lift your school bags upwards by 1 m What is NOT considered as work done: –You push against a wall –Jumping continuously on the same spot –Holding a chair and walking around the classroom

5. 5 Physics concept of WORK Homework: Page 261 – 262 #’s 1-8 all page 278 #’s 54, 61,63,65 WORK can be calculated by: W = F x d Units: [J] [N] [m] SI Unit for Work is JOULE (J) Lesson 1 Feb 10

6. 6 Examples of WORK You are helping to push Romac’s heavy shopping cart with a force of 50.0 N for 200.0 m. What is amount of work done? Work done, W = F x d = 50.0 x 200.0 = 10,000 J or 10.0 kJ (kilo-Joules) Lesson 1 Feb 10

7. 7 Examples of WORK: Thomas put on his bag-pack of weight 120 N. He then starts running on level ground for 100 m before he started to climb up a ladder up a height of 10 m. How much work was done? From Physics point of view, no work is done on pack at level ground. Reason: Lift is perpendicular to movement. Work is done on pack only when Thomas climbs up the ladder. Work done,W = F x s = 120 x 10 = 1200 J or 1.2 kJ Lesson 1 Feb 10

8. 8 Lesson 2 Feb 13 Mechanical Advantage Objectives Lesson 2 MA Objectives: –Show understanding of the Physics concept of Mechanical Advantage –Show understanding of the Physics concept of IDEAL Mechanical Advantage –Correctly identify MA and IMA from given situations –Recall and show understanding of the formula to calculate Efficiency of a System –Do NOW: If you push a 10.0 Kg object from rest along a frictionless surface with a force of 2.0 N what is the work done on the object in the first 5.00 seconds? ANSWER DUE @ 10:27 SHARP – Lesson 2 HOMEWORK: –Glencoe PAGE:280-281: #’s 79, 80, 81, 82, 85, 86, (87 – 88)

9. What is a Simple Machine? A simple machine has few or no moving parts. Simple machines make “work” easier Do NOW: If you Lift a 10.0Kg mass 1.0 meter off the floor how much work have you done? 9 Lesson 2 HOMEWORK Lesson 2 HOMEWORK : Glencoe PAGE:280-281: #’s 79, 80, 81, 82, 85, 86, (87 – 88) Glencoe Page 273 #29, 30, 32

10. Write this DOWN Ideal Mechanical Advantage {IMA}is the RATIO of the –Displacement exerted (in) to the Displacement load (out). IMA = d in / d out mechanical advantage Apply the concept of mechanical advantage to everyday situations. N Conservation of Energy –Work in = Work out F in * d in = F out * d out Mechanical Advantage {MA} is the RATIO of the –Force exerted (in) to the Force load (out). MA = F out / F in Efficiency (%) is the RATIO of the (MA) / (IMA) * 100 10 Lesson 2

11. Practice leverIf you lift a 10.0Kg mass 1.0 meter off the floor with a lever. The Fulcrum is.50 meters from the mass and 3.50 meters from the other end. If it takes a 2.0 Kg mass to balance the lever, What is the IMA of the Lever? –IMA = Dist in / Dist out = 3.5 / 0.5 = 7:1 What is the MA of the Lever? –MA = Force out / Force in –10.0*9.8 / 2.0 *9.8 = 98 / 19.6 = 5:1 What is the efficiency of the Lever? Efficiency = MA / IMA * 100 = 5 / 7 *(100) = 71.4286 %Efficiency = MA / IMA * 100 = 5 / 7 *(100) = 71.4286 % 11 Lesson 2

12. PE  KE Grade Homework Glencoe Page 271 Example 4 HOMEWORK –Lesson 2 HOMEWORK: –Glencoe PAGE:280-281: #’s 79, 80, 81, 82, 85, 86, (87 – 88) Homework –Glencoe Page 272 #’s 24 – 28 Homework 12 Lesson 2 – A FEB 16 IN CLASS Page 273 #’s 29, 30, 32

13. 13 LESSON 3 KINETIC ENERGY Feb 16 th DO NOW: What is the acceleration of a 4,500 Kg auto as it goes from 29.5 m/s to zero in 10.0 meters? At the end of the lesson, students should be able to: –Understand the Physics concept of Kinetic Energy (KE) = ½ m v 2 –Recall and show understanding of the KE formula –Distinguish situations involving KE –Demonstrate knowledge of the Work Energy Theorem – Work = ΔKE Lesson 3 HOMEWORK: PAGE:287: #’s 1 – 3 Page : 291: #’s 4 – 8 Page: 297 #’s 15 – 18

14. 14 Energy – Quick Re-cap Energy is the capacity to do work SI Unit: Joule (J) Many forms Common ones: –Kinetic –Potential –Electric –Chemical –Solar –Nuclear LESSON 3

15. 15 Kinetic Energy (KE) Formula: KE = ½ mv 2Formula: KE = ½ mv 2 The amount of KE of a moving body depends on: –Mass of body (kg) –Velocity (m/s) –When the Mass doubles : KE Doubles QUADRUPLES –When the Velocity doubles: KE QUADRUPLES SI Unit: Joule [ J ] … same as Work Done Work = Δ KE = ½ m (Vf – Vi) 2 LESSON 3

16. 16 Examples of KE Find the KE of an empty van of mass 1000kg moving at 2m/s. Find the KE of van when it is loaded with goods to give a total mass of 2000kg, and moving at 2m/s. Find KE of unloaded van when it speeds up to 4m/s. KE of van at 2m/s = ½ x 1000 x (2) 2 = 2000 J = 2 kJ KE of van at 2m/s = ½ x 2000 x (2) 2 = 4000 J = 4 kJ KE of van at 2m/s = ½ x 1000 x (4) 2 = 8000 J = 8 kJ LESSON 3

17. 17 Examples of KE A motorcycle accelerates at 2m/s 2 from rest for 5s. Find the KE of motorcycle after 5s. Mass of motorcycle is 200 kg. Velocity of motorcycle after 5s,a = (v-u) t v = 2(5) + 0 = 10m/s KE of motorcycle at 10m/s= ½ x 200 x (10) 2 = 10,000 J = 10 kJ LESSON 3

18. 18 Lesson 4 Feb 17 Potential Energy DO NOW: What is the work done on a 10.0 Kg book lifted to a table 1.5 meters from the floor? If Potential Energy is defined as m*g*h what is the Potential Energy of that same 10.0 Kg book when it has been raised to the table? At the end of the lesson, you should be able to: –Show understanding of the Physics concept of Gravitational Potential Energy PE = mgh –Recall and understand the formula PE = mgh –Distinguish situations involving GPE –Solve related problems involving GPE HOMEWORK: Page: 308 #’s 64 – 68 all

19. 19 Potential Energy Potential energy is the energy possessed by an object as a result of its POSITION or CONDITION. Two common kinds: –Gravitational PE  GPE –Elastic PE (not in syllabus) In Physics, ground level is normally assumed to be at ZERO GPE. Any object that is at ground level has ZERO GPE. If object is lifted a certain height above ground, its GPE has increased. Lesson 4 Feb 17

20. 20 Gravitational PE Can be calculated with: GPE = mass  gravitational  height above acceleration ground level = m  g  h Units: [J] [kg] [m/s 2 ] [m] SI Units of GPE : Joule [J] Ground, 0 GPE Distance from ground, h Object on top of building, of mass, m g earth Lesson 4 Feb 17

21. 21 Example of GPE You lifted your bags to the top of your table. What can you say about the GPE of your bag? –Zero, increase, decrease Lift the same bag on the Moon. What happens to GPE? –Zero, increase, decrease Will the GPE be the same on Earth and Moon? –Same, less on Moon, more on Moon? Lesson 4 Feb 17

22. 22 Examples of GPE You lifted a set of books of mass 3kg, for 2m. What is the GPE gained by the books? Take g=10m/s 2. Find the work done by you to lift the books. GPE = mgh = 3  10  2 = 60 J Work done, W = F  d (F = weight of books) = (m  g)  d = 3 x 10 x 2 = 60 J (Note: same as GPE) Lesson 4 Feb 17

23. Conservation of Energy At the end of the lesson, students should be able to:  Show understanding of conservation & conversion of energy  Correctly distinguish situation involving energy conservation & conversion  Solve related problems 23 Lesson 5 Feb 20 Energy of an object can be thought of as the sands in an hourglass! CANNOT Note that energy CANNOT be created nor destroyed! Energy always remain same or fixed in quantity! But this sand can change position, from the top to bottom and bottom to top! Likewise energy can change in form eg. From KE  PE Do Now: What is the PE of a 5.0 Kg mass raised 4.0 meters above the ground? If all that energy is converted to Kinetic Energy after it fall 4.0 meters what will the Velocity be when it hits the ground? Use TWO Methods: mgh = ½ mv 2 Vf 2 = Vi 2 + 2ad EP – TL WK - WS ZB – PL BV - AL TP – BZ CM - HH

24. http://hyperphysics.phy- astr.gsu.edu/hbase/hframe.html 24 http://hyperphysics.phy- astr.gsu.edu/hbase/flobj.html#c1 Conservation & Conversion of Energy HOMEWORK LESSON 5 Page: 308 – 309 #’s 71, 73, 78, 80 Lesson 5 Feb 20

25.  Conversion of energy is the term used to denote change in energy from one form to another.  Eg.  Burning candle: Chemical  Heat, Light  Fuel: Chemical  Heat  KE  Electricity  Nuclear explosion: Nuclear  Heat, light  Spring: Elastic PE  KE 25 Lesson 6 Feb 21 More Conservation of Energy TE = PE + KE – Work {friction} DO NOW: A) Turn in last nights homework B) What is the velocity of a 2000.0 Kg roller coaster at the bottom of a 42.0 m hill? C) What is the velocity ½ way down the hill? HOMEWORK: Page 309 #’s 81 – 89 ODD TEST

26. More of Conservation of Energy  A fresh Coconut of mass 5 kg is found growing at the end of a tree branch 20 m above ground. When ripe, the Coconut will by itself drops to the ground below. Let gravity = 10m/s 2.  Find the energy of the fresh coconut? What form is it?  GPE. GPE = mgh = 5 x 10 x 20 = 1000J  Find the GPE and KE of the coconut when it is 5m above ground. Sum up both the GPE and KE and compare the value with above. What can you infer from the results?  GPE = 5 x 10 x 5 = 250J.s = ½ vt, v = gt s = ½ gt 2, t = √ 3 KE = ½ mv 2 = ½ (5)(10 √ 3) 2 = 750Jv = 10(√ 3)  Sum of energies = 250 + 750 = 1000J  Same as above => energy is conserved. 26 Lesson 6 Feb 21

27. More Conversion of Energy  A car of 800 kg is moving at an average speed of 5 m/s. The traffic light changed to red and so the driver stepped on the brakes to bring the car to a quick, sudden and screeching halt.  Find energy of moving car and what form of energy is this?  KE. KE = ½ mv 2 = ½ x 800 x 5 2 = 10,000 J.  What energy does the car possesses when it stops?  None.  What happened to the original energy of the moving car?  KE has changed to Sound and Heat Energy. 27 Lesson 6 Feb 21

28. Work – Energy Review 28 Lesson 7 Feb 22