# Sol-Ch05 - CHAPTER 5 E xercises 1 F x = T 1 c os 40 T 2 c...

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Exercises 1.  ∑ F x = T 1 cos 40° – T 2 cos 60° = 0 F y = T 1 sin 40° + T 2 sin 60° – 7g = 0 Find T 1 = 34.8 N ; T 2 = 53.4 N. 2. F x = T 1 cos 30° – T 2 = 0 ; F y = T 1 sin 30° – 3g = 0 T 1 = 58.8 N, T 2 = 50.9 N 3.  ∑ F x = F – T sin 37° = 0; F y = T cos 37° – mg = 0. (a) F = 14.7N; (b) T = 24.5 N 4.  ∑ F y = 2T cos θ – mg = 0, where cos θ = 1.5 / 50, thus T = 11.4 kN 5. F 1 + F 2 = ma , thus –i + 2 j +3k + F 2 = 8i – 6 j   , and F 2 = 9i –8 j – 3k N. 6. F 1 = i + 2 j N and F 2 = 3.19i + 2.41 j N, a = (F 1 + F 2 ) / (0.2 kg) = 21.0i + 22.1 j   m / s 2 7. a = (4i j ) / (1.5 kg) = 2.67i – 0.67 j   m / s 2 v = v 0 + a t = (2i + 3 j ) + (5.34i – 1.34 j ) = 7.34i + 1.66 j m / s. 8. a = (400) 2 / (0.06 m) = 2.67 × 10 6 m / s 2 . F = ma = 2.67 × 10 4 N, about 45 times the person’s weight 9. (a) v 2 = v 0 2 + 2aΔx, so a = (6 ×  10 13 ) / (0.04) = 1.5 × 10 15 m / s 2 . F = m a = 1.37 × 10 15 N (b) t = (6 ×  10 6 ) / (1.5 × 10 15 ) = 4 × 10 9 s. 10. a = v 2 / 2Δx = 1/6 = 0.167 m / s 2 . ΣF x = mg sin θ – f = ma , so f = 109 N 11. (a) = v 2 / 2Δx, thus a = 22.5 m / s 2 ; (b) f = (50 kg) a = 1125 N. 12. (a) a = 2.67 m / s 2 , F = ma = 3.27 kN; (b) a = –7.56 m / s 2 , F = ma = –9.26 kN In both cases, F is the frictional force exerted by the road. 13. (a) a = 31.6 m / s 2 , so F = 3.95 × 10 5 N (b) a = –31.25 m / s 2 , so F = –3.90 × 10 5 N. CHAPTER 5

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14. F – mg = ma, so F = m(g + a) = (0.215 kg) (13.8 m / s 2 ) = 2.97 N 15. To rise by 0.4 m, the initial speed v required is given by 0 = v 2 – 19.6 × 0.4, so v 2 = 7.84 m 2 /s 2 . This v is the final speed while the feet are in contact with the ground, thus v 2 = 2a (0.15) and a = 26.1 m/s 2 is the acceleration of the torso. The force exerted is F – mg = ma, so F =1795 N. 16. (a) F x = F cos θ – mg sin θ = ma, so a = –1.90 m / s 2 (down); (b) Δ x = (4)(2) – 1/2 (1.9)(2 2 ) = +4.2 m (up incline) 17. With v 0 = 4.17 m / s, 0 = (4.17) 2 – 2(g sin 10°) Δ x, so Δ x = 5.1 m. 18. (a) Fcos θ – f = 0, f = 69.3 N; (b) F’cos θ – f = ma, F’ = 103 N 19. From x = 1/2 at 2 , a = 14 m / s 2 and F = ma = 840 N. 20. v y = a y t = (F / m)t = –40 m / s (S). Thus v = 10i – 40 j   m / s or 41.2. m/s at 76° S of E. Path is parabolic. 21. (a) a = v
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Sol-Ch05 - CHAPTER 5 E xercises 1 F x = T 1 c os 40 T 2 c...

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