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Work

Work. Work is the product of the ………………applied in the direction of motion, and the ………………of the object. W = … x …. The unit of work is ……… called a Joule (J). One joule is the amount of work done when a force of …N acts over …. m. 1J = ……………. = 1………… -2 .m = 1kg.m 2 .s -2

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Work

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  1. Work Work is the product of the ………………applied in the direction of motion, and the ………………of the object. W = … x …. • The unit of work is ……… called a Joule (J). • One joule is the amount of work done when a force of …N acts over …. m. 1J = ……………. = 1…………-2.m = 1kg.m2.s-2 • Work is a ………… quantity. (…………….. Not necessary) • From the formula we see that if there is no ………………….. then there is no work done. • If the displacement is …………………………… to the force then there is no work done either. F

  2. Work Work is the product of the resultant force applied in the direction of motion, and the displacement of the object. W = F•xcos • The unit of work is N.m called a Joule(J). • The joule is the amount of work done when a force of 1N acts over 1m. 1J = 1N.1m = 1kg.m.s-2.m = 1kg.m2.s-2 • Work is a scalar quantity. ( Direction not necessary) • From the formula we see that if there is no movement then there is no work done. • If the displacement is perpendicular to the force then there is no work done either. F  x

  3. Energy – Kinetic Energy Energy of movement W = f…. F = m…. =>W = ……… vf2 = vi2 +2a x but vi = 0 => x = vf2 ..….. W = ……………… => EK = ………………... • Energy is the capacity to do ………….. • Energy is a scalar quantity and has the same unit as work viz. ………... • When ENERGY is transferred …………. is done. • If an object is accelerated from rest. FORCE m m a m.s-2 m m

  4. Energy – Kinetic Energy Energy of movement W = f.x F = m.a =>W = max vf2 = vi2+2ax but vi = 0 => x = vf2 /2a W = m.a.(vf2/2a) => EK = 1/2mvf2 • 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. FORCE m m a m.s-2 m m

  5. Potential energy Object lifted to a height “h”. Energy gained (or transferred) = ………………. Ep of an object = ………….. to lift the object to that height W = ……… F = ……… W = …………. Ep = …………. Where h is the displacement of lifted object. h

  6. Potential energy Object lifted to a height “h”. Energy gained (or transferred) = work done Ep of an object = work done to lift the object to that height W = f.x Fg = mg W = mgx Ep = mgh Where h is the displacement of lifted object. h

  7. Potential energy Bouncing Efficiency Task: Investigate the efficiency of the bounce of a ball. Theory: The efficiency of the bounce can be expressed as the % of energy of the bounce over the original potential energy of the ball. Efficiency = x100 Outcomes: Write up all the steps of the scientific method and include: Calculations of energy and efficiency from at least five different heights. Also show at least one calculation of the velocity of the ball on impact with the surface. Compare the efficiency of at least two different balls. h initial Initial Potential energy (top) Potential energy of the bounce h bounce

  8. Mechanical Energy An object dropped from REST EK = ____ Ep = . Ep = ____ Ek = . • Mechanical energy is the SUM of the and energy of an object at any given time. E = . • The energy, which is converted or transferred, is equal to the to do this. • If an object is moving and is brought to rest then the to do this is equal to the . • At any point during the fall of an object the is equal to the potential energy of the object it began to fall.

  9. Mechanical Energy EK = 0 Ep = mgh Ep = 0 Ek = (mgh) • Mechanical energy is the sum of the kinetic and potential energy of an object at any given time. E = Ep + E`k • The energy, which is converted or transferred, is equal to the work done to do this. • If an object is moving and is brought to rest then the work done to do this is equal to the change in kinetic energy. Ek = work done • 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. • If any energy is lost through work then E = Ep + Ek + Work done

  10. Mechanical Energy A 4kg ball is dropped 8m. E = Ep + Ek Any object that is above the ground has the POTENTIAL to fall Down. The potential energy EP =mgh(Height) lost would beconverted into velocity (kineticenergy) EK. Ep(……) = Ek(……….) If the object is lifted up then it ………..potential energy. At any time the …….of these two is called the MECHANICAL ENERGY - and it stays ………………. 4kg m Ep = …… Ek = ….J E = Ep + Ek = …+ … = ……J Ep Potential Energy LOST 8m E = Ep + Ek = … + …. = 320J 4m EK kinetic energy ………..= Ep ……! E = Ep + Ek = … + … = …..J Ep = … m Ek = ……

  11. Mechanical Energy 4kg m Any object that is above the ground has the POTENTIAL to fall Down. The potential energy EP =mgh(Height) lost would beconverted into speed (kineticenergy) EK. Ep(Top) = Ek(Bottom) If the object is lifted up then it gainspotential energy. At any time the sum of these two is called the MECHANICAL ENERGY - and it stays constant. E = Ep + Ek = 320 + 0 = 320J Ep = mgh Ek = 0J Ep Potential Energy LOST 8m E = Ep + Ek = 160 + 160 = 320J 4m EK kinetic energy gained = Ep lost! E = Ep + Ek = 0 + 320 = 320J Ep = 0 m Ek = 320J E = Ep + Ek

  12. Conservation of energy - Pendulum Energy cannot be created or destroyed but merely transferred from one form to another. Whenever there is energy conversion there is ………… done. A C h B The …………of the pendulum is not required to calculate the velocity. Ep (Top) = Ek (Bottom) ………… = ………….2 => V = √………..

  13. Conservation of energy - Pendulum Energy cannot be created or destroyed but merely transferred from one form to another. Whenever there is energy conversion there is work done. A C h B The mass of the pendulum is not required to calculate the velocity. Ep (Top) = Ek (Bottom) Mgh = 1/2mv2 => V = √2gh

  14. A pendulum question • 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 = (60)(10)(30) + 0 = 18000J m h EBottom= Ep + Ek = mgh + 1/2mv2 18 000 = 0 + 1/2 (60)v2 v2 = 18000/30 = 600 v = 600 = 24.5 m.s-1

  15. 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 = (60)(10)(30) + 0 = 18000J m h EBottom = Ep + Ek = mgh + 1/2mv2 18 000 = 0 + 1/2 (60)v2 v2 = 18000/30 = 600 v = 600 = 24.5 m.s-1

  16. Power • Power is the ………………….. at which WORK is done. Power = Units: Watts (W) = Since v = ……….., Power can be found by P = …………. If a force of 20N is exerted over a distance of 5m for a time of 30s the power used would be.

  17. Power Work time • Power is the RATE at which WORK is done. Power = Units: Watts (W) = Joules (J) Time (s) Since v = s/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

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