1 / 18

Work, Energy

Work, Energy. &. Power. Review. ( Joules = N · m ). ( J ). Work = Force x distance. Example : How much work is done by a crane that lifts 120 kg up 75 m?. F = mg. W = F·d. = m·g·d. = 120 x 9.8 x 75. (note the distance here is the height). = 88000 J.

jag
Download Presentation

Work, Energy

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. Work, Energy & Power

  2. Review (Joules = N·m) (J) Work = Force x distance Example:How much work is done by a crane that lifts 120 kg up 75 m? F = mg W = F·d = m·g·d = 120 x 9.8 x 75 (note the distance here is the height) = 88000 J Power = rate of doing work Power = rate of using energy Power = Work / time (Watts = J/s) (W)

  3. Example:Find the power output of a 60 kg athlete who climbs the Grouse Grind in 40 minutes. (elev. gain = 800m) Power = Worktime = (mg)d t = 60 x 9.8 x 800 2400 = 196 W (in seconds)

  4. Gravitational Potential Energy The work done to lift (or lower) something is equal to its change in gravitational potential energy e.g.: Potential Energy (PE)(at height “h”)= work done to lift the object to that height. m h h=0 (reference point) PE = F·d = m·g·h PE = m·g·h

  5. Example:Find the potential energy of a 60 kg student sitting on a 80 cm high stool… • relative to the floor in the room. • relative to the floor in the room 1 floor below (3.0 m lower) a) PE = mgh = 60 x 9.8 x 0.8 = 470 Joules b) PE = mgh = 60 x 9.8 x 3.8 = 2200 Joules

  6. Energy & Work An object is accelerated from initial velocity, vo, to velocity, v, by a force, F (net force), over a distance, d.What is the work done? F = ma F·d = ma·d F·d = m(a·d) 2ad = v2– vo2 F·d = m(v2 - vo2)2 W = ½mv2 - ½mvo2 Kinetic Energy (energy of motion) KE = ½mv2 W = KE - KEo W = ΔKE (Work – Energy Theorem)

  7. Work – Energy Theorem W = ΔKE The work done on an object by the net force equals the change in kinetic energy of that object. Example:A 1000 kg car going 20 m/s brakes and comes to a stop. How much work is done? What happens to the energy? KE = ½mvo2 = ½(1000)(20)2 = 200000 J W = ΔKE = 0 – 200000 J = –200000 J (lost energy) Energy is dissipated as heat.

  8. Homework • Practice Problems: Pg. 221 (1, 2) • Practice Problems: Pg. 224 (5, 7) • Pg. 236 R.C. (5, 7) A.C. (1, 3) • Pg. 236 Problems (2, 4, 5, 10)

  9. Conservation of Mechanical Energy Total Mechanical Energy of an object moving in a gravitational field: E = PE + KE E = mgh + ½mv2 Conservation Law:If no other force (than gravity) does any work on an object in a gravitational field, the total mechanical energy is conserved. Eo = Efinal

  10. Example: • A soccer ball of mass 0.5 kg is kicked with an initial velocity of 15 m/s. Find its… • velocity when it is 4.0 m high. • height when it is going 5.0 m/s. v h 15m/s a) Eo = Efinal mgho + ½mvo2 = mgh + ½mv2 0 + ½(0.5)(15)2 = (0.5)(9.8)(4.0) + ½(0.5)v2 56.3 = 19.6 + 0.25v2 v = 12 m/s b) 56.3 = (0.5)(9.8)h+ ½(0.5)(5.0)2 h = 10 m

  11. Tarzan, a 75 kg Ape-man, swings from a branch 3.0 m above the ground with an initial speed of 5.0 m/s. • Find his velocity when he swings past ground level. • Find the maximum height he swings to. • What should his initial speed be if he is to just reach a branch 4.0 m high? b) & c) 3.0 m a) h = 0 Note: in the physics jungle there is no air resistance and the vines have no mass and are unstretchable!

  12. h = vo2 + ho2g Note: The vine does NO WORK because it is always applying its force perpendicular to the direction of motion a) mgho + ½mvo2 = mgh + ½mv2 m cancels out; h = 0 gho + ½vo2 = 0 + ½v2 v = √(vo2 + 2gho) v = √((5.0)2 + 2(9.8)(3.0)) v = 9.2 m/s horizontal b) gho + ½vo2 = 0 + gh v = 0 h = 4.3 m c) gho + ½vo2 = 0 + gh v = 0 vo = √(2gh - 2gho) vo = 4.4 m/s

  13. Tarzan’s less famous brother, George, swings from the same 3.0 m-high branch with an initial speed of 3.5 m/s. Unfortunately, he hits a tree 1.5 m above the ground. Find his speed just before impact and the work done by the tree. His mass is 90 kg. 3.0 m 1.5 m

  14. mgho + ½mvo2 = mgh + ½mv2 m cancels out gho + ½vo2 = gh + ½v2 v = √(vo2 + 2gho – 2gh) v = √((3.5)2 + 2(9.8)(3.0 – 1.5)) v = 6.5 m/s W = ΔKE = KE - KEo W = ½mv2–½mvo2 W = 0 – ½(90)(6.5)2 W = -1900 J

  15. Efficiency Definition: Efficiency = useful output x 100% input Example: • A crane operates at 15 kW and lifts a 700 kg weight 50 m in 2 minutes. Find the efficiency. Power = Energy / time Input: Ei = P x t =15000W x 120s =1.8 x 106 J Ouput: Eo = PE =mgh =700 x 9.8 x 50 =3.43 x 105 J Efficiency: Eff =3.43 x 105x 100%1.8 x 106 =19%

  16. Eff =EoEi =Eo0.15 =32340.15 • A 6.5 hp forklift is 15% efficient when lifting a 220 kg box. How long does it take to lift the box 1.5m? (1 hp = 746 W) Ouput: Eo =mgh =220 x 9.8 x 1.5 =3234 J Efficiency: =0.15 Input: Ei =21560 J Ei = P x t 21560 = (6.5 x 746) x t t=4.5s

  17. =½mv2 x 100%P x t 0.65 =½ (800) (27.8)2(100 x 746) t Eff =Eox 100%Ei 0.65 =309000 74600 x t • An 800 kg, 100 hp car is 65% efficient at transferring energy to the wheels. How long does it take to go from 0  100 km/h? Efficiency: t=6.4s

  18. Homework

More Related