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CHAPTER 5 Work and Energy

Work: Work done by an agent exerting a constant force is defined as the product of the component of the force in the direction of resulting displacement and the magnitude of the displacement. W = F cos  · d. CHAPTER 5 Work and Energy. Dimensional Analysis. W = F · d = N · m

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CHAPTER 5 Work and Energy

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  1. Work: Work done by an agent exerting a constant force is defined as the product of the component of the force in the direction of resulting displacement and the magnitude of the displacement. W = F cos · d CHAPTER 5Work and Energy

  2. Dimensional Analysis W = F · d = N · m = kg · m/s2 · m = kg · m2/s2 = Joule 1 N · m = 1 kg · m2/s2 = 1 Joule

  3. Fa m d mv2 = F · d 2 KE = mv2 2 Energy of Motion = Kinetic Energy = mv2 2 Energy: The means to do work Energy also has the units of “Joules”. What happened to the energy that caused the force that did the work? Assume: Frictionless Surface FNet,x = F = ma a = F/m Work is transformed into energy of motion. Kinematics: vo = 0m/s v = ? a = F/m t = ? x = d v2 = 2 ·F/m · d

  4. F d=h Fg = mg Assume the object is moved at constant velocity upward. W = F · d W = mgh Work is transformed into energy of elevation. Energy of Elevation = Potential Energy = mgh PE = mgh

  5. Work Energy Equation In the absence of friction; work, kinetic energy, and potential energy are conserved. Win + KE1 + PE1 = KE2 + PE2 Win = Work put into the object or done on the object KE1 & PE1 = Kinetic and Potential Energy of the object before the work was performed. KE2 & PE2 = Kinetic and Potential Energy of the object at the end of the event. Fa · d + ½ mv12 + mgh1 = ½ mv22 + mgh2

  6. Example Problem (Frictionless Rollercoaster) A motor does work as a 250kg coaster car is raised vertically 85 meters up an incline. How much work is “done on” the car? W = F · d = mgh = 250kg · 9.80m/s2 · 85m = 210,000 kg ·m2/s2 = 210,000 Joules 210,000J of work resulted in 210,000 of Potential Energy

  7. Example Problem (Frictionless Rollercoaster) The 250 kg car is at rest on the top of the incline when it is released and travels to the bottom of the track. What is its velocity at the bottom? Win + KE1 + PE1 = KE2 + PE2 0 + 0 + 210,000J = KE2 + 0 210,000J = ½ mv22 210,000J = ½ (250kg)v22 v2 = 41m/s Loss of 210,000J of PE results in gain of 210,000J of KE

  8. Principle of Conservation of Energy: • Energy can never be created or destroyed. • Energy may be transformed from one form into another. • The total energy in an isolated system remains constant. • The total energy of the universe is constant. • Mother of All Energy Equations: • Win + KE1 + PE1g = Wout + KE2 + PE2g • Win = Fa · d • Wout = Ff · d • Wout generally dissipates to heat which also has units of Joules.

  9. Power is the rate at which work is done P = W t Dimensional Analysis P = W = Joules = Watt t sec P = W = 8100J t 10.sec Power 1 Watt = 1 Joule/sec Example Problem (Power) A 55kg rock is lifted at constant velocity 15m in 10.sec by a motor. What is the power output of the motor? Work = PE = mgh = 8100 Joules t = 10.sec P = 810J/s P = 810 Watts

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