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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.

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Work, Energy

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  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

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