1. Unit 3 Work, Energy & Power Serway Chapters 7 & 8 Glencoe Chapter 10 & 11 1 Unit 3 Section 3 Conservation of Energy

2.  Unit 3 Section 3 Lesson 1 Conservation of Energy  Conservation of Energy Objectives  Show understanding of conservation & conversion of energy  Correctly distinguish situation involving energy  Conservation & Conversion  Solve related PE and KE problems  Take into account FRICTION  Total Energy = PE + KE - Friction  DO NOW  What is the potential energy of a roller coaster car with a total mass of 2500.00 kg that is at the top of a 65.0 m (measured from the ground) hill?  HOMEWORK: Glencoe Page 308 #’s: 73 – 83 Odd  Physicslab Conservation of Energy: http://dev.physicslab.org/Document.aspx?doctype=5&filename=C ompilations_CPworkbook_ConservationEnergy.xml http://dev.physicslab.org/Document.aspx?doctype=5&filename=C ompilations_CPworkbook_ConservationEnergy.xml 2 Unit 3 Section 3 Lesson 1 Conservation of Energy

3. Conservation of Energy  Energy of an object can be thought of as the sands in an hourglass!  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  eg. From KE  Friction (heat) 3

4. Conservation vs. Conversion of Energy  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 4

5. Conversion of Energy  Eg. of KE  PE  Roller-coaster  Falling object 5

6. Free Falling object model  An object in free fall means the object is falling freely, under the influence of gravity 6 When the object is at the highest position, the GPE is at maximum and KE is zero. When the object is falling, the GPE decreases as it loses height, and the KE increases At the lowest position, the KE is at maximum and GPE is zero. GPE lost = KE gained

7. IN CLASS Conservation of Energy  A fresh durian of mass 5 kg is found growing at the end of a tree branch 20 m above ground. When ripe, the durian will by itself drop to the ground below. Let gravity = 10m/s 2.  Find the energy of the fresh durian just before it drops? What form is it?  GPE = mgh = 5 x 10 x 20 = 1000JGPE  Find the GPE and KE of the durian 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.  Vf 2 = Vi 2 + 2gd  0 +2(10)15 = 300  Vf 2 = 300  V = 17.321m/s  KE = ½ mv 2 = ½ (5)(17.321) 2 = 750 J  OR ….KE = GPE original – GPE 5m = 750 J  Sum of energies = PE:250 + KE:750 = 1000J  Same as above => energy is conserved. 7

8. 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. 8

9. Conservation of Energy  Energy cannot be made or destroyed but it can change form, total energy stays constant  Consider a marble at the top of a ramp with 2.5 J of energy… GPE GPE = 2.5 J KE KE = 0 J GPE GPE = 0 J KE KE = 2.5 J GPE GPE = 1.25J KE KE = 1.25 J Total Energy- 2.5 J 2.5 J2.5 J a b c cb a Ignoring heat energy losses!

10. More Example Questions AA stone of mass 3 kg is dropped from a height of 60 cm onto the ground. a) Calculate its GPE before it is dropped GGPE = Weight x height = mgh = (3 x 10) x 0.6 = 18 J b) When it hits the ground how much GPE does it have? How much KE does it have? GPE is zero as its height off the ground is zero. KE is 18 J, all the GPE has been converted into KE c) At what speed will it hit the ground? KE = ½ mv 2 18 = (½) (3)(v 2 ) 12 = v 2 3.46 m/s = v

11.  Unit 3 Section 3 Lesson 2 Conservation of Energy  Conservation of Energy Objectives  Solve related PE and KE problems  Take into account FRICTION  Total Energy = PE + KE - Friction  Potential Energy Curves  DO NOW  What is the potential energy of a roller coaster car with a total mass of 2500.00 kg that is at the top of a 65.0 m (measured from the ground) hill and neglecting friction, how fast will it be traveling at the bottom of the hill if the track has a 55 degree incline from the horizontal and is 79.35 meters long?  HOMEWORK: Glencoe Page 309 #’s: 84– 93 All 11 Unit 3 Section 3 Lesson 2 Conservation of Energy

12. FRICTION and WORK and ENERGY  Total Energy usually = PE at start  PE initial + KE initial = PE final + KE final – F friction *d  F friction = μ mg COS{ Ɵ }ramp  Or F friction = μ m(v 2 /r) curve 12

13. Examples  A ball of mass 2 kg is dropped from a height of 5.0 m above the floor. Find the velocity of the ball as it strikes the floor?  E 0 = KE i + PE i = KE f + PE f  0 + mgh = ½ mv 2 + 0  0 + 2(10)(5.0 ) = ½ (2) v 2  v = √ (100) = 10m/s  A box is pushed UP an inclined plane with an angle of 37 degrees with an initial velocity of 10.0 m/s. If the surface is frictionless, how high up the ramp will the box go when it stops?  E 0 = KE i + PE i = KE f + PE f  ½ mv 2 + 0 = 0 + mgh  ½ 10 2 + 0 = 0 + gh  h = 50/10 = 5 m  How far up the ramp {distance} will it slide?  Sin Ɵ = h / d  d = h / Sin Ɵ =8.3 m 13 h d Ɵ

14.  A skier starts from rest at the top of a 20.0 º incline and skis in a straight line to the bottom of the slope over a distance of 400.0 m. If the coefficient of friction between the skis and the snow is 0.2, calculate the skier’s speed at the bottom of the hill.  E 0 = KE i + PE i – 1 Work friction = KE f + PE f  E 0 = 0 + mgh – μ mgcos ( Ɵ ) * d = ½ mv 2 + 0  g( d * Sin Ɵ ) – μ gcos ( Ɵ ) * d = ½ v 2  V = √{2g( d * Sin Ɵ ) – 2 μ gcos ( Ɵ ) * d } = 35 m/s 14 Friction Examples h d Ɵ

15. Potential Energy Curves  E 0 = KE i + PE i – 1 Work friction = KE f + PE f  E 0 = KE i + U(x) i – 1 Work friction = KE f + U(x) f   v = + √ { (2/m) [E 0 – U(x)]}  F(x) = dU/dx  Example Hooks Law  F(x) = d/dx (1/2 kx 2 ) = kx 15 Potential Energy Curves and Questions PHYSLAB