Work W = Fd = _____

1. DET Work W = Fd = _____ W W Fd Fd KE due to ___________ of entire object Q within molecular and atomic _______ PE stored up due to ___________ motion bonds position mechanical energy SYSTEM PE + KE + Q Total energy of a system: ET =________________ Work W = Fd = DET = D( ) =

3. DET Work W = Fd = _____ W W Fd Fd KE due to ___________ of entire object Q within molecular and atomic _______ PE stored up due to ___________ motion bonds position mechanical energy SYSTEM PE + KE + Q Total energy of a system: ET =________________ Work W = Fd = DET = D( ) = PE + KE + Q DPE + DKE + DQ

4. gravitational: DPE = mgDh DPE = elastic: DPE = Dposition ½ kx2 motion of entire object: DKE = acceleration Work W = Fd = DET ½ mv2 friction heat DQ _________  D____________ energy  DKE of atoms/molec. or DPE or ________  increase_________. or change__________ internal bonds temp phase

5. What kind of energy does the work change if you…. 1/ …lift an object at a constant speed?  constant v  no D_____  ____________________ DPE or D___ KE gravitational Q 2/ …accelerate an object at a constant height?  constant h  no D____  D_____ or D____ PE KE Q 3/ …drag an object across a surface at a constant speed and height?  constant v  no D______  constant h  no D______  D___ only  energy from work against___________ goes into raising __________ or changing _________ KE PE friction Q phase temp.

6. KE due to motion of entire object Q within molecular and atomic bonds PE stored up due to position mechanical energy Work done on a system DET However, if NO work is done on a system, then its total energy ET cannot change—it stays the same. In other words, get rid of the "external work," and the total energy ET will stay the same forever. X X Fd Fd This is what we call an isolated system.

7. Conservation The Law of the _______________________ of Energy: The total energy of an isolated system of bodies remains _________________. constant after before = ET = PE + KE + Q = ET' PE' + KE' + Q' total The __________ energy of a system is neither increased nor decreased in any process. Energy cannot be __________ or _______________ . It can only be_______________ (changed) from one type to another. created destroyed transformed Q' 0 If there is no friction, DQ = ____  Q = ____ . Then: = PE' + KE ' PE + KE mechanical mechanical _____________ energy before ______________ energy afterward

8. Notice the similarity between: …the Conservation of Energy: PE + KE = PE' + KE' …and the Conservation of Linear Momentum: p1 + p2 = p1' + p2' • Momentum conservation is used to determine what • happens to objects after they collide or explode apart. • Energy conservation is also used to determine what • happens to objects at a later time. • Both momentumandenergyare useful ideas • because they allow us to ignore forces, which can • be very complicated.

9. Restatements of the… Law of the Conservation of Energy: "The total energy is neither __________________ nor __________________________ in any process." "Energy can be __________________ from one form to another and _______________ from one body to another, but the __________ amount remains ________________." "Energy is neither _____________ nor _______________." increased decreased transformed transferred total the same destroyed created

10. Ex: A pendulum swings back and forth. It position at two points is shown below. Ignore friction. What energy does it have at each position? Use: ET = ET' 0 At max. height, v = ____ so KE will = ______ 0 When it reaches maximum height: PE = 20 J (given) KE =_____ ET = When it reaches minimum height: PE' = ? KE' = ? ET' = ? 0 J 0 20 J 20 J 20 J PEtop=> KEbottom Notice:

11. Ex: A 0.25-kg box is released from rest and slides down a frictionless incline. Find its speed as it arrives at the bottom of the incline. ET = ET' Use: 0.25 kg 4.0 m When the box is released at the top: PE = mgh = KE = because vi = ____ ET = = 9.81 J (0.25 kg)(9.81m/s2)(4.0 m) 0 0 9.81 J

12. Just before it hits bottom: PE' = ? KE' = ? ET ' = ? 0 (h = 0) 9.81 J 9.81 J KEbottom PEtop => Now use the equation for KE to find v: KE = (1/2) mv2 9.81 J = (0.5) (0.25 kg)v2 78.5 = v2 8.9 m/s = v

13. Ex: Drop a 2.0 kg rock off of a 4.0-m cliff. Use g ≈10 m/s2 to simplify. Do not ignore air resistance. (2.0 kg)(10m/s2)(4.0 m) = 80 J mgh = PE = KE =_____ ET = 0 80 J 4.0 m Just before it hits: PE = Suppose KE = 75 J (PEtop ≠ KEbottom) ET = 75 J + ???? 0 How much energy is "missing?" And where did it go? How will it affect v at bottom? 5 J air resistance Q  heat v will be less

14. Ex: The mass m is not really needed to find the speed v when using the Conservation of Energy with no friction: KEbottom PEtop => top h v = ? bottom ET (top) ET (bottom) PE + KE PE' + KE' PE + 0 0 + KE' mgh (1/2)mv2 gh (1/2)v2 v (2gh)1/2  indep. of m!!! = = = = = =

15. Ex: A spring with a spring constant of 220 N/m is compressed a distance 0.035 m as shown below. A mass of 0.027 kg is place against it on a frictionless slide. When the mass is released: 1/ how fast will it go as it leaves the spring, and 2/ how high up the slide will it go before it stops and comes back down? slide m energy stored energy of mass energy of mass in spring as it starts moving at highest point   = = KE PEs PEg (1/2)mv2 mgh (1/2)kx2

16. 1/ How fast will the mass be going as it leaves the spring? = = = = = KE PEs (1/2)mv2 (1/2)kx2 (1/2)(0.027 kg)v2 (1/2)(220 N/m)(0.035 m)2 0.135 J (0.0135)v2 v 3.2 m/s 2/ How high up the slide will it go? = = = = = PEg PEs (1/2)kx2 mgh 0.135 J (0.027 kg)(9.8 m/s2)h 0.135 J (0.265) h h 0.51 m

17. Ex. Mr. Siudy is fired out of a cannon at 3 different angles with the same speed from a cliff. 1/ For which angle will he hit the ground with the most speed? 2/ For which angle will he hit the ground in the least time? same All 3 cases begin with the: ________ PE _________KE _________ET same 1 same 2 3 When he reaches the bottom in all 3 cases, he will have the: same ________ PE _________KE _________ET same v same same 3 In case _____ he reaches the ground in the least time.