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Solution to Pairs Exercise #1

Solution to Pairs Exercise #1. +. M=4000 kg. 20 m. +. +. Datum. Solution to Pairs Exercise #1. D E p =Net Mech. Work Input Net Mech. Work Input = D E p = mgh = (4000 kg)(9.8 m/s 2 )(20 m)(1 J·s 2 /(kg·m 2 )) = 784,000 J. Solution to Pairs Exercise #2.

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Solution to Pairs Exercise #1

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  1. Solution to Pairs Exercise #1 + M=4000 kg 20 m + + Datum

  2. Solution to Pairs Exercise #1 DEp=Net Mech. Work Input Net Mech. Work Input = DEp = mgh = (4000 kg)(9.8 m/s2)(20 m)(1 J·s2/(kg·m2)) = 784,000 J

  3. Solution to Pairs Exercise #2 h1= 0.35 (coal to electricity) (see Example 21.24) h2= 0.03 (Figure 21.30) hoverall = h1 h2 = (0.35)(0.03)=0.0105

  4. Solution to Bungee #1 Total Energy = Ep + Ek = mgh + ½ kx2 + ½ mv2 Energy on bridge = Energy at Xtaut mg(0) + ½ k(0)2 + ½ m(0)2 = 75 kg (9.81 m/s2)(-50.0 m) + ½ k(0)2 + ½ (75 kg) v2 v = 31.3 m/s

  5. Solution to Bungee #2 Total Energy = Ep + Ek = mgh + ½ kx2 + ½ mv2 Energy at bridge = Energy at hmax mg(0) + ½ k(0)2 + ½ m(0)2 = (75 kg)(9.81 m/s2)(-hmax) + ½ (15 kg/s2)(hmax – 50 m)2 + ½ m(0)2 0 = -735.75 hmax + 7.5 (hmax)2 - 750 hmax + 18750 0 = 7.5 (hmax)2 – 1485.75 hmax + 18750 hmax = 184.6 m or 13.6 m V = 0 at hmax Stretched length only Only valid solution

  6. Solution to Bungee #3 Total Energy = Ep + Ek = mgh + ½ kx2 + ½ mv2 Energy at bridge = Energy at river mmaxg(0) + ½ k(0)2 + ½ mmax (0)2 = mmaxg xriver + ½ k(xriver – xtaut)2 + ½ mmaxv2 0 = mmax (9.81 m/s2)(-195 m) + ½ (15 kg/s2) (195 m – 50 m)2 + ½ mmax(0)2 mmax = 82.4 kg

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