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AP Physics Unit 3 Work , Energy & Power

AP Physics Unit 3 Work , Energy & Power. Unit 3 Section 1 Work and Mechanical Advantage. Serway Chapters 7 Glencoe Chapter 10. Unit 3 Section 1 Lesson 1 Work and the Work-Energy Theorem Work Objectives. Unit 3 Section 1 Lesson 1 Work Objectives :

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AP Physics Unit 3 Work , Energy & Power

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  1. AP Physics Unit 3 Work, Energy & Power Unit 3 Section 1 Work and Mechanical Advantage Serway Chapters 7 Glencoe Chapter 10

  2. Unit 3 Section 1 Lesson 1Work and the Work-Energy TheoremWork Objectives Unit 3 Section 1 Lesson 1 Work Objectives: • Show understanding of the Physics concept of Work • Correctly identify Work from given situations • Recall and show understanding of the formula to calculate Workdone • Solve related problems involving Work • Do NOW: If you push a 10.0 Kg object from rest along a frictionless surface with a force of 2.0 N what will the velocity be in 5.00 seconds? • Unit 3 Section 1 Lesson 1 HOMEWORK: • Serway PAGE:207: #’S 3,4,5,7,11,12,13, 15, 17, 20, 22

  3. In Class DO THIS in memory of ME • Review Vectors • F = Fi + Fj + Fk i = x j = y k = z • d = di + dj + dk • (F)•(d) = (Fd)x + (Fd)y + (Fd)z • d = √(di2 + dj2 + dk2) • F = √(Fi2 + Fj2 + Fk2) • Review Integrals (Power rule) • (Hooks Spring law)  F = kx • W = ∫ifkx dx = ½ k(xfinal2) – ½ k(xinitial2) – • Serway page 188 Examples 7.2 - 7.3 -7.4 - 7.5 - 7.6 • Serway PAGE:206 : #’S 1 – 5, 12 • Serway PAGE:207: #’S 3, 4, 5, 7, 11, 12, 13, 15, 17, 20, 22

  4. 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. • NO MOTION  NO WORK

  5. 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

  6. Physics concept of WORK WORK can be calculated by: Work done = Constant x Distance moved {in the direction of motion}force (N) in the direction of force (m) W = (F)(d)(Cos{Ɵ} ) Units: [J] [N] [m] SI Unit for Work is JOULE (J)

  7. Scalar (“dot”) Product of two VectorsConstant Force and Displacement W = (F)•(d) = (F)(d)(Cos{Ɵ}) F Ɵ d Integralproduct of two VectorsNon-ConstantForce and Displacement W = ∫Fx dx

  8. More Examples of WORK • You are helping to push your mother’s heavy shopping cart with a force of 50 N for 200 m. What is amount of work done? Work done,W = (F)(d)(Cos{Ɵ} ) = (50)(200)(1) = 10,000 J or 10 kJ (kilo-Joules)

  9. More Examples of WORK: • Jack 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 Jack climbs up the ladder. Work done,W = (F)(d)(Cos{Ɵ} ) • = (120)(10)(1) • = 1200 J or 1.2 kJ

  10. Unit 3 Section 1 Lesson 2 JAN 27Mechanical AdvantageObjectives Unit 3 Section 1 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? • Unit 3 Section 2 Lesson 2 HOMEWORK: • Glencoe PAGE:280-281: #’s 79, 80, 81, 82, 85, 86, (87 – 88)

  11. In Class • Glencoe Page 268 Challenge Problem • Glencoe Page 271 Example 4 • Glencoe Page 272 #’s 26 – 28 • Unit 3 Section 2 Lesson 2 HOMEWORK: • Glencoe PAGE:280-281: #’s 79, 80, 81, 82, 85, 86, (87 – 88)

  12. What is a Simple Machine? • A simple machine has few or no moving parts. • Simple machines make “work” easier

  13. Mechanical Advantage 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

  14. Labs • Lab 3-1:1 • Work done on a Flexible Flyer Sled • (inside- a cart) • Lab 3-1:2 • Build a simple machine

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