Sol-Ch08 - CHAPTER 8 E xercises 1 1/2(m 1 m 2 v 2 m 1 g h m 2 g h = 0 T hus v = 1.83 m/s 2 1/2(m 1 m 2 v 2 m 1 g h = 0 so v = 1.71 m/s 3 1/2(m 1 m

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Exercises 1. 1/2(m 1 + m 2 ) v 2 – m 1 g h + m 2 g h = 0. Thus, v = 1.83 m/s 2. 1/2 (m 1 +m 2 ) v 2 – m 1 g h = 0, so v = 1.71 m/s 3. 1/2 (m 1 + m 2 ) v 2 – m 2 g d sin θ + m 1 g d sin θ = 0, so v = 1.18 m/s 4. ΔU = mg h = 2.97 × 10 5 J. One gram of fat provides 5.64 kJ of mechanical energy. Thus, ΔU requires 52.7 g of fat. 5. With the given values E = 1/2mv 2 + mg L (1 – cos θ ) = 1.79 J (a) 1/2 mv 2 = 1.79 J, so v = 2.44 m/s (b) mg L(1 – cos θ 0 ) = 1.79, so θ 0 = 53.6° 6. E 1 = mgL 1 (1 – cos θ 1 ) = 1.47 J; E 2 = mgL 2 (1 – cos θ 2 ) = 1.47 J, thus θ 2 = 50° 7. (a) 1/2 kA 2 = 0.8 J = 1/2 mv max 2 , so v max = 2.53 m/s; (b) 1/2 kx 2 + 1/2 mv 2 = 0.8 ; so v = 2.19 m/s (c) U = E/2, i.e., 1/2 kx 2 = 1/4 kA 2 ; thus x = 28.3 cm 8. (a) 60 J; (b) –60 J; (c)ΔK = –ΔU, so v f = 8.72 m/s 9. mg (x + 0.6) = 1/2 kx 2 , 60 x 2 – 4.9x – 2.94 = 0, so x =26.6 cm 10. E 1 = mgL = 6.125 J, E f = mgL (1 – cos θ ) + 1/2 mv 2 F y = T = mgcos θ = mv 2 /L. Using T = 6 N mv 2 = L (6 – mg cos θ ). Substitute mv 2 into E f = 6.125 J to find θ = 65.9° 11. (a) –mg D + 1/2 kD 2 = 0, so D = 1.96 m (b) –mg(0.5) + 0.5(6)v 2 + 0.5(40)(0.25) = 0, so v = 2.21 m/s 12. (a) k = 4.9 N/m, E i = 1/2 kA 2 = 1/2(4.9)(0.3) 2 = 0.22 J E f = mg h =0.22, so h =44.9 cm (b) E i = 1/2 kA 2 , E f = 1/2 kx 2 + mg (A + x) = 0.22 J, Thus, 2.45x 2 + 0.49x – 0.073 = 0, so x = 0.1 m 13. (a) 1/2 kD 2 – m 1 gD + m 2 gD = 0, D =1.23 m CHAPTER 8
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(b) 0.5(m 1 + m 2 ) v 2 + (m 2 – m 1 ) g d + 0.5kd 2 = 0; v = 0.95 m/s 14. (a) 1/2 mv 2 = mg d sin θ , so v = 6.26 m/s ; (b) E i = 1/2 mv 2 = 1.96 J, E f = 1/2 kA 2 – mgA sin θ . Thus, 2.5A 2 – 0.49A – 1.96 = 0, so A = 0.99 m 15. (a) E 1 = 0.5mv 2 + mgy 1 = 165 J; U 2 = 157 J. It gets over hill. (b) E 3 = 1/2 kA 2 + mgy 3 = 165, so A = 1.31 m 16. (a) mg L = 1/2 mv max 2 . F y = T – mg = mv max 2 /L, so T = 3mg = 23.5 N (b) mg L = 1/2 mv 2 + mg L (1 – cos θ ),   F y = T – mg cos θ = mv 2 /L, thus T = 2mg cos θ = 18.8 N. 17. E = mgH = 3U/4 + U = 7mgy/4; thus y = 4H/7. 18. (a) U = E/2, thus mgy = 1/2(1/2 mv i 2 ), so y = 40.8 m; (b) U = 2E/3, thus mgy = 2/3(1/2 mv i 2 ), so y = 54.4 m 19. E A = 1/2 mv A 2 + mgy A = 2.19 × 10 5 J. (a) E
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This note was uploaded on 11/09/2011 for the course EE EE taught by Professor Ee during the Spring '11 term at National Chiao Tung University.

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Sol-Ch08 - CHAPTER 8 E xercises 1 1/2(m 1 m 2 v 2 m 1 g h m 2 g h = 0 T hus v = 1.83 m/s 2 1/2(m 1 m 2 v 2 m 1 g h = 0 so v = 1.71 m/s 3 1/2(m 1 m

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