# Sol-Ch06 - CHAPTER 6 E xercises 1. ( a) From v 2 = v 0 2 +...

This preview shows pages 1–3. Sign up to view the full content.

Exercises 1. (a) From v 2 = v 0 2 + 2aΔx, with a = –f/m, we find f = 0.135 N. (b) f = µmg, so µ= 0.153 2. (a) f = µmg/2 = ma. Thus, a = µg/2 = 3.92 m/s 2 ; (b) All four wheels contribute, so f = µmg, a = µg = 7.84 m/s 2 3. (a) 5g sin θ – T – µ(5g cos θ ) = 0, T – µ(2g) = 0. Adding these : 5g sin θ = µ(5g cos θ + 2g). Thus, µ= 0.395 (b) T = 2µg = 7.74 N. 4. (a)  ∑ F x = mg sin θ µ(mg cos θ ). Since µ s > tan θ , block does not start to move. (b) ∑ F y = mg cos θ – F sin θ ; and f = µN. F x = F cos θ – mg sin θ µ k (mg cos θ + F sin θ ) = ma, a = 6.86 m/s 2 5. (a) µ s Mg – µ k mg = ma; leads to a = 27.4 m/s 2 (b) Need the force exerted by the person on the crate : F = µ s Mg = 627 N 6. (a) Maximum f s = µ s mg = 34.3 N. Thus, f = F = 30 N; (b) 30 + µ k mg = ma, so a = 10.9 m/s 2 (to the right) ; (c) 30 – µ k mg = ma, so a = 1.1 m/s 2 (to the right) 7. (a) Fcos37° – µ s (mg + F sin37°) = 20 – 0.5(44.4) < 0, so it does not start to move. (b) 20 – 0.2(44.4) = 3a so a = 3.71 m/s 2 . 8. (a) Fcos37° – µ s (mg – F sin θ ) = 20 – 0.5(14.4) > 0, so it does move. (b) 20 – 0.2(14.4) = 3a, so a = 5.71 m/s 2 9. (a)  ∑ F x = mg sin θ µ s (mg cos θ ) > 0, so the body does move. F x = 19.6 – µ k (14.7) = 2.5a, so a = 6.37 m/s 2 , downward (b) –mg sin θ µ k (mg cos θ ) = ma; thus a = 9.31 m/s 2 (downward) CHAPTER 6

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
(c) mg sin θ µ k mg cos θ = ma; thus a = 6.37 m/s 2 (downward) 10. f = ma = µmg, so µ = a/g = 0.612 11. Given v 0 = 27.8, from 0 =v 0 2 – 2aΔx, we find a = μ g = 6.43 m/s 2 , so μ =0.656. (a) g sin θ μ (g cos θ ) = a = –4.63 m/s 2 , so Δ x = 83.3 m. (b) g sin θ + μ (g cos θ ) = a = 8.03 m/s 2 , so Δ x = 48.1 m. 12. (a) mg sin θ μ (mg cos θ ) = ma, so a = 5.55 m/s 2 ; (b) v = v 0 + at, so t = 4.0 s; (c) Δ x = v 2 /2a = 44.5 m 13. a = g sin θ + μ g cos θ = 2.67 m/s 2 . From E q. 3.12, 0 = (22.2) 2 – 2aΔx, so Δ x = 92.3 m. 13 (a) t R / t F = 1.5; (b) Δ x R / Δ x F = 1 15. (a) F – μ (m A + m B + m C )g = 100a ; a = 1.02 m/s 2 (b) F – μ m A g – T 1 = m A a, thus T 1 = 140 N. (c) T 2 μ m C g = m C a, so T 2 = 40 N. 16. x = 1/2 at 2 yields a = 0.533 m/s 2 mg sin 20 – μ (mg cos θ ) = ma, so μ = 0.306 17. F x = F cos θ μ (mg + F sin θ ) = F(cos θ μ sin θ ) – μ mg (a) Need F > μ s mg/ (cos θ μ sin θ ) for block to move. (b) If μ s = cot θ , F = ; (c) a = – μ k g 18. (a)  ∑ F x = F cos θ μ (mg cos θ + F sin θ ) – mg sin θ = ma, 20.0 – 0.1(39.1 + 15.0) – 29.5 = 5a, so a = –2.98 m/s 2 (downward) (b) Δ x = 6 × 2 – 1/2(2.98) (2 2 ) = 6.04 m 19. (a) 0; (b) For m A : f = ma gives μ (2g) = 2a, so a = 2.45 m/s 2 . The maximum force on system is F = 7a = 17.2 N.
This is the end of the preview. Sign up to access the rest of the document.

## 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.

### Page1 / 9

Sol-Ch06 - CHAPTER 6 E xercises 1. ( a) From v 2 = v 0 2 +...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online