Published on May 14, 2010
Work Energy & Power W = F • ∆x cosθ F) rce( Fo θ Fx ∆x Keith Warne
Energy – Kinetic Energy Energy is the capacity to do work. Energy of movement W = f.∆x Energy is a scalar quantity and has the F = m.a same unit as work viz. Joule. =>W = ma∆x When ENERGY is v f 2 = vi2 +2a∆x but transferred WORK is vi = 0 and vf = v done.A mass (m) is accelerated from restto a velocity v. => ∆x = v 2 /2a Vi = 0 vf = v W = m.a.(v 2/2a)FORCE m m m m.s -2 a m SAMPLE ONLY => EK = 1/2mv 2
Work Energy Theorem The work done on an object by a net force is equal to the change in the object’s kinetic energy: EK = 0 ∆Ek = work done (Net) Ep = mgh If V = const then ∆Ek = W nett = ........ With Conservative forces (...............................) at any point during the fall of an object the mechanical energy is equal to the potential energy of the object before it began to fall. ∆E (mech) = ......... E p (........) = E k (.............) Ek = (mgh) Ep = 0 With Non conservative forces present (friction): Energy is lost through work then E =SAMPLE+ Ek + ........................... Ep ONLY SAMPLE ONLY
An object is lifted vertically by a force F at a constant velocity. F WF = F.∆ x cos θ 2 kg = (- Fg).∆ x cos θ = -(2*-9.8)*(3)*(1)= 58.8 J 3m Wg = F.∆ x cos θ Fg = Fg.∆ x cos (180) = (2*9.8)*(3)*(-1)= -58.8 J Work Energy Theorem ∆Ek = work done (Net) = 0 (const v) Work done by F + Work done by g = 0 W f = -W g .: F nett = 0 .: F + F g = 0 .: F = -F gSAMPLE ONLY SAMPLE ONLY
An object is pushed up a slope by a force F at constant velocity. Frictionless slope. 3m F 2 kg H = 3sin30 Fg 30o Fnett = 0 (const v) .: |F // | = |F| F // = F g sin 30 WF = F.∆ x cos θ = 2*9.8 sin 30 = F// ∆ x cos θ = (2*9.8*sin30)*(3)*cos(0) = 29.4 JWg = F.∆ x cos θ = (2*9.8)sin30 *(3)cos (180) = - 29.4 JWg = F.∆ x cos θ = Fg .H cos (180) = (2*9.8)*3*sin(30)*(-1) = - 29.4 J Work Energy Theorem (W nett ) SAMPLE ONLY W nett = ∆SAMPLE ONLY v.) Ek = 0 (const
A pendulum • A person of 60kg is lifted to a height of 30m on a slingshot pendulum and then released what is its maximum speed?Total Mechanical Energy(Top)ETop = Ep + EK = mgh + 1/2mv2 m = (60)(10)(30) + 0 = 18000J h EBottom = Ep + Ek = mgh + 1/2mv2 18 000 = 0 + 1/2 (60)v2 SAMPLE ONLY v2 SAMPLE ONLY 600 = 18000/30 = -1
Power • Power is the RATE at which WORK is done. Work Power = time Joules (J) Units: Watts (W) = Time (s) Since v = ∆x/t, Power can be found by P = F.vIf a force of 20N is exerted over a distance of 5m for a time of 30s the powerused would be. W = F.∆x = (20).5 = 100J P = W/t = 100/30 = 3.3 W SAMPLE ONLY SAMPLE ONLY
WE Th Eg2 m=2kg f=2N F=4N S 10 m D• The horizontal force F of 4 N is pulling the block whilst a frictional force f of 2 N opposes the motion. If the block’s velocity is 2 m.s-1 when it passes point S, what is its velocity when it passes point D? How much work is done against friction?Fnet = F + f .: Fnett = 4 – 2 = 2 N W f = Ff∆xcosθ = 2*10*cos(180)and the angle between Fnet and ∆x is 0o = 20*(-1)Wnet = ∆Ek Fnet∆xcosθ = ∆Ek = -20 JF∆xcos0o = ½ mvf2 – ½ mvi22x10x1 = ½ 2vf2 – ½ 2x22 = vf2 – 4vf2 = 20 + 4 = 24∴vf = ONLYSAMPLE4.90 m.s-1 SAMPLE ONLY
a) Work done by 680N force: Note on Work WF = F.Δxcosθ W = 680 x 80 x cos0 = 54400 Jb) The work done on the object by the gravitational force. 680 N Fg = mg = 40 x 9,8 = 392 N Wg = F.Δxcosθ = 392 x 80 x cos180 = -31360 J 40kgc) The net work done on the object. Wnet = WF + Wg = 54400 – 31360 = 23040 Jd) The gain in potential energy U = mgh = 40 x 9.8 x 80 = 31360 Je) The gain in kinetic energy. 392 N Wnet = ΔK = 23040 J 80m• The speed of the rocket. K = ½ mv2 23040 = ½ x 40 x v2 v2 = 23040/20 v = 33,9 m.s-1 SAMPLE ONLY SAMPLE ONLY speed = 33,9 m.s-1
Keith Warne• This is a SAMPLE presentation only• The full presentations contain loads more slides (20-100) and are available on the click2learn website here• http://www.click2learn.co.za/shop/lessoncloud-grade-12-science-slides/ (paste into your browser if link above does not work)• I have other resources available from my Science Café site feel free to have a lookSAMPLE ONLY SAMPLE ONLY
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