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# Work, Energy & Power

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A set of slides created to teach Work, Energy & Power to learners at Bishops Diocesan College in Cape Town.

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### Transcript of "Work, Energy & Power"

1. 1. For FULL presentation click HERE >> www.warnescience.net Work Energy & Power Keith Warne x  Fx W = F • x cos
2. 2. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net m mm Energy – Kinetic Energy  Energy is the capacity to do work.  Energy is a scalar quantity and has the same unit as work viz. Joule.  When ENERGY is transferred WORK is done. Energy of movement W = f.x F = m.a =>W = max vf 2 = vi 2 +2ax but vi = 0 and vf = v => x = v 2 /2a W = m.a.(v 2/2a) => EK = 1/2mv 2m a m.s-2FORCE A mass (m) is accelerated from rest to a velocity v. Vi = 0 vf = v m
3. 3. For FULL presentation click HERE >> www.warnescience.net 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 = work done (Net) If V = const then Ek = Wnett = ........  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) = ......... Ep (........) = Ek (.............)  With Non conservative forces present (friction):  Energy is lost through work then E = Ep + Ek + ........................... Wnett = Wf1 + Wf2 + etc. EK = 0 Ep = mgh Ek = (mgh) Ep = 0 SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY
4. 4. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net An object is lifted vertically by a force F at a constant velocity. 2 kg 3 m F WF = F.x cos  = (- Fg).x cos  = -(2*-9.8)*(3)*(1)= 58.8 J Wg = F.x cos  = 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 Wf = -Wg .: Fnett = 0 .: F + Fg = 0 .: F = -Fg Fg SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY
5. 5. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net An object is pushed up a slope by a force F at constant velocity. Frictionless slope. 3 m Fnett = 0 (const v) .: |F//| = |F| WF = F.x cos  = F// x cos  = (2*9.8*sin30)*(3)*cos(0) = 29.4 J Wg = F.x cos  = (2*9.8)sin30 *(3)cos (180) = - 29.4 J Wg = F.x cos  = Fg .H cos (180) = (2*9.8)*3*sin(30)*(-1) = - 29.4 J Work Energy Theorem (Wnett ) Wnett = Ek = 0 (const v.) 30o Fg F// = Fg sin 30 = 2*9.8 sin 30 H = 3sin30
6. 6. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY 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? m h ETop = Ep + EK = mgh + 1/2mv2 = (60)(10)(30) + 0 = 18000J Total Mechanical Energy (Top) EBottom = Ep + Ek = mgh + 1/2mv2 18 000 = 0 + 1/2 (60)v2 v2 = 18000/30 = 600 v = 600 = 24.5 m.s-1 SAMPLE ONLY
7. 7. For FULL presentation click HERE >> www.warnescience.net Power • Power is the RATE at which WORK is done. Power = Units: Watts (W) = Work time Joules (J) Time (s) Since v = x/t, Power can be found by P = F.v If a force of 20N is exerted over a distance of 5m for a time of 30s the power used would be. W = F.x = (20).5 = 100J P = W/t = 100/30 = 3.3 W SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY
8. 8. For FULL presentation click HERE >> www.warnescience.net WE Th Eg2 • 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 and the angle between Fnet and ∆x is 0o Wnet = ∆Ek Fnet∆xcos = ∆Ek F∆xcos0o = ½ mvf 2 – ½ mvi 2 2x10x1 = ½ 2vf 2 – ½ 2x22 = vf 2 – 4 vf 2 = 20 + 4 = 24 vf = 4.90 m.s-1 m=2kg F = 4 N S 10 m D f = 2 N Wf = Ff∆xcos = 2*10*cos(180) = 20*(-1) = -20 J SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY
9. 9. For FULL presentation click HERE >> www.warnescience.net Note on Worka) Work done by 680N force: WF = F.Δxcosθ W = 680 x 80 x cos0 = 54400 J b) The work done on the object by the gravitational force. Fg = mg = 40 x 9,8 = 392 N Wg = F.Δxcosθ = 392 x 80 x cos180 = -31360 J c) The net work done on the object. Wnet = WF + Wg = 54400 – 31360 = 23040 J d) The gain in potential energy U = mgh = 40 x 9.8 x 80 = 31360 J e) The gain in kinetic energy. Wnet = ΔK = 23040 J f) The speed of the rocket. K = ½ mv2 23040 = ½ x 40 x v2 v2 = 23040/20 v = 33,9 m.s-1 speed = 33,9 m.s-1 40kg 80m 680 N 392 N SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY
10. 10. For FULL presentation click HERE >> www.warnescience.net Hi - This is a SAMPLE presentation only. My FULL presentations, which contain loads more slides and other resources, are freely available on my resource sharing website: www.sciencecafe.org.za (paste into your browser if link above does not work) Have a look and enjoy! Keith Warne
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