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Work, Energy & Power

Work, Energy & Power. Chapter 10. 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

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Work, Energy & Power

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  1. Work, Energy & Power Chapter 10

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

  6. Examples of WORK Lesson 1 Feb 10 • 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)

  7. Examples of WORK: Lesson 1 Feb 10 • 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

  8. Lesson 2 Feb 13Mechanical AdvantageObjectives 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. Lesson 2 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? • 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 Lesson 2 Apply the concept of mechanical advantage to everyday situations. • Ideal Mechanical Advantage {IMA}is the RATIO of the • Displacement exerted (in) to the Displacement load (out). • IMA = d in / d out • Conservation of Energy • Work in = Work out • F in * d in = F out * d out N • 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

  11. Practice Lesson 2 • If 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 %

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

  13. LESSON 3KINETIC ENERGY Feb 16th 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 v2 • 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. Energy – Quick Re-cap LESSON 3 • Energy is the capacity to do work • SI Unit: Joule (J) • Many forms • Common ones: • Kinetic • Potential • Electric • Chemical • Solar • Nuclear

  15. Kinetic Energy (KE) LESSON 3 • Formula: KE = ½ mv2 • The amount of KE of a moving body depends on: • Mass of body (kg) • Velocity (m/s) • When the Mass doubles : KE Doubles • When the Velocity doubles: KE QUADRUPLES SI Unit: Joule [ J ] … same as Work Done Work = Δ KE = ½ m (Vf – Vi)2

  16. Examples of KE LESSON 3 • 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

  17. Examples of KE LESSON 3 • A motorcycle accelerates at 2m/s2 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

  18. Lesson 4 Feb 17Potential 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 • Recall and understand the formula PE = mgh • Distinguish situations involving GPE • Solve related problems involving GPE HOMEWORK: Page: 308 #’s 64 – 68 all

  19. Potential Energy Lesson 4 Feb 17 • 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.

  20. Object on top of building, of mass, m g earth Distance from ground, h Ground, 0 GPE Gravitational PE Lesson 4 Feb 17 • Can be calculated with: GPE = mass  gravitational  height above acceleration ground level = m  g  h Units: [J] [kg] [m/s2] [m] SI Units of GPE : Joule [J]

  21. Example of GPE Lesson 4 Feb 17 • 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?

  22. Examples of GPE Lesson 4 Feb 17 • You lifted a set of books of mass 3kg, for 2m. What is the GPE gained by the books? Take g=10m/s2. • 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)

  23. Lesson 5 Feb 20 EP – TL WK - WS ZB – PL BV - AL TP – BZ CM - HH Conservation of Energy 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 mv2 Vf2 = Vi2 + 2ad 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 Energy of an object can be thought of as the sands in an hourglass! 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 formeg. From KE  PE

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

  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? • 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 HOMEWORK: Page 309 #’s 81 – 89 ODD TEST

  26. Lesson 6 Feb 21 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/s2. • 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 gt2, t = √ 3 KE = ½ mv2 = ½ (5)(10 √ 3)2 = 750J v = 10(√ 3) • Sum of energies = 250 + 750 = 1000J • Same as above => energy is conserved.

  27. Lesson 6 Feb 21 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 mv2 = ½ x 800 x 52 = 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.

  28. Lesson 7 Feb 22 Work – Energy Review

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