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This document features detailed solutions to various exercises related to mechanical work and energy transformations. Solutions include net mechanical work calculations for mass and height scenarios, energy conservation principles in bungee cord applications, and conversions from coal to electricity efficiency metrics. Key exercises break down the total energy equations, highlighting potential energy, kinetic energy, and elastic potential energy components. The content serves students and educators in physics to enhance understanding of energy concepts and mechanical systems.
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Solution to Pairs Exercise #1 + M=4000 kg 20 m + + Datum
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
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
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
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
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
