1 / 17

Work W = Fd = _____

D E T. 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.

palmer
Download Presentation

Work W = Fd = _____

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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( ) =

  2. 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

  3. 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

  4. 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.

  5. 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.

  6. 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

  7. 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.

  8. 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:

  9. 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

  10. 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

  11. 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

  12. 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!!! = = = = = =

  13. 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

  14. 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 Nm)(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

  15. Ex. Mr. Butchko 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.

  16. Open your Review Book packet to pages: 88-89 Do problems #67-80

More Related