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CalculatingEnergy Contents: Gravitational Potential Whiteboards Kinetic Whiteboards

CalculatingEnergy Contents: Gravitational Potential Whiteboards Kinetic Whiteboards. Gravitational Potential Energy. Lifting a box of mass m: W = Fs = energy given F = mg, s =  h W = Fs = mg  h. m.  h. There are two ways to lift the object.  E p = mg  h

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CalculatingEnergy Contents: Gravitational Potential Whiteboards Kinetic Whiteboards

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  1. CalculatingEnergy • Contents: • Gravitational Potential • Whiteboards • Kinetic • Whiteboards

  2. Gravitational Potential Energy Lifting a box of mass m: W = Fs = energy given F = mg, s = h W = Fs = mgh m h There are two ways to lift the object... • Ep = mgh • Ep - gravitational potential energy • h - Change in height • m - Mass • g - 9.81 N/kg on Earth TOC

  3. Whiteboards: Gravitational Potential Energy 1 | 2 | 3 TOC

  4. What is the potential energy of a 4.5 kg bowling ball, 13.5 cm above the ground? Ep = mgh, h = .135 m Ep = (4.5 kg)(9.81 N/kg)(.135 m) = 5.959575 (6.0 J) W 6.0 J

  5. Toby Continued lifts a 75.0 kg box doing 1573 J of work. What is the change in height of the box? Ep = mgh h = Ep/(mg) = (1573 J)/(75 kg)/(9.81 N/kg) = 2.13795… W 2.14 m

  6. Colin Host lifts himself up 15 m doing 9555 J of work. What is his mass? Ep = mgh m = Ep/(gh) = (9555 J)/(9.81 N/kg)/(15 m) = 64.9337… W 65 kg

  7. Kinetic Energy v F d m Speeding up a box of mass m with force F W = Fs = energy given F = ma, s = d W = (ma)d v2 = u2 + 2as, d = v2/(2a) (if u = 0) W = (ma)(v2/(2a)) = mv2/2 = 1/2mv2 Ek = 1/2mv2 • Ek = 1/2mv2 • Ek - Kinetic energy • v- velocity • m - mass There are two ways to speed up the mass... TOC

  8. Whiteboards: Kinetic Energy 1 | 2 | 3 TOC

  9. What is the kinetic energy of a 4.20 g bullet going 965 m/s? Ek = 1/2mv2 = 1/2(.0042 kg)(965 m/s)2 = 1955.5725 (1960 J) W 1960 J

  10. A European swallow has 2.055 J of kinetic energy when it is flying at 14.23 m/s. What is its mass in grams? Ek = 1/2mv2 m = 2 Ek/v2 = 2(2.055 J)/(14.23 m/s)2 = .020297… (20.29 g) W 20.29 g

  11. What speed must a .563 kg hammer move to store 34 J of energy? Ek = 1/2mv2 v = (2 Ek/m) = (2(34 J)/(.563 kg)) = 11 m/s W 11 m/s

  12. A 4.0 kg shot is sped up from 6.0 m/s to 9.0 m/s. What is the change in kinetic energy? initial: Ek = 1/2mv2 = 1/2(4.0 kg)(6.0 m/s)2 = 72 J final: Ek = 1/2mv2 = 1/2(4.0 kg)(9.0 m/s)2 = 162 J So the increase is 162 – 72 = 90. J increase W 90. J

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