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# Sol-Ch07 - CHAPTER 7 E xercises 1 W = Fs cos = 2 4 J 2 W =...

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Exercises 1. W = Fs cos θ = 24 J 2. W = (2i – 3 j   + k ) (2i – 2 j   –4k ) = 6 J 3. 40 runs of 10 m = 400 m, thus W = (40N) (400 m) = 16 kJ. 4. (a) W P = Fs = 240 J; (b) W g = –mg Δ y = –147 J; (c) W f = –f s = –66 J 5. W = f s = (µN) (12 ×   2πr) = (8 N) (12) (2πr) = 24.1 J 6. ΣF x = Fcos θ – f = 0,   F y = N + Fsin θ – mg = 0. Thus, Fcos = f = µ(mg – Fsin θ ) or F = µmg/(cos θ + µsin θ ) = 4.99 N (a) W F = Fs cos θ = 7.06 J (b) W f = –fs = –7.06 J; (c) W g = 0 7. F x = Fcos θ – f = 0, F y = N – Fsin θ – mg = 0. Using f = µN, find F = µmg / (cos θ – µsin θ ) = 8.32 N. (a) W f = Fs cos θ = 11.8 J ; (b) W f = –fs = –11.8 J ; (c) W g = 0. 8. W g = –mg Δ y = –29.4 J is independent of the path taken 9. (a) W g = –mg Δ y = +mgs sin θ = 49.7 kJ; (b) W f = –fs = –4 kJ. 10. F = 157.5 N, then W F = mg Δ y = 1.89 kJ 11. (a) W motor = –mgs sin θ = –134 J; (b) +134 J 12. W = (mg) (v Δ t) = 784 J CHAPTER 7

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13. (a) Net area = +30 J (Displacement and force are positive) ; (b) Area = –10 J (Displacement is negative, force is positive) 14. (a) +F 0 A/2 ; (b) –F 0 A/2 15. (a) +F 0 A/2 ; (b) –F 0 A/2 16. (a) –F 0 A/2 ; (b) F 0 A/2 17. (a) W ext = 1/2 k(x f 2 – x i 2) = 0.2 J; (b) W ext = 0.6 J 18. k = ΔF/Δ x = 73.5 kN/m (Assume weight shared equally by all springs) 19. W ext = 1/2 kx 2 . (a) w 1 /w 2 = k 1 /k 2 ; (b) W = 1/2 k(F/k) 2 α 1/k. Thus, W 1 /W 2 = k 2 /k 1 20. ΔK = 1/2 m(v f 2 – v i 2 ) = 1(29 – 13) = 16 J 21. K = 1/2 mv 2 = 1/2 m(2πr/T) 2 , where T = 1 y and r = 1.5 × 10 11 m. Find K = 2.66 × 10 33 J = 6.3 × 10 23 tons 22. K = 1/2 mv 2 , W = FΔ x = ΔK, so Δ x = ΔK/F (a) 120 J, 0.15 m; (b) 26.2 kJ, 3.28 m; (c) 3.62 MJ, 4.53 km; (d) 3.48 × 10 10 J, 4.36 ×   10 7 m 23. (a) ΔK = –1/2 (0.2) (400) = –40 J; (b) W g = –mg Δ y = –35.3 J; (c) No, because of air resistance 24. K = 1/2 mv 2 , W = FΔ x = ΔK, so F = ΔK/Δ x.
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