29.1. Appendix A: Answers to Selected Problems (3e) http://www.ck12.org
1. 1450 N
2. 5600 N
3. 5700 N
- Friction between the tires and the ground
- Fuel, engine, or equal and opposite reaction
- (b) 210 N (c) no, the box is flat so the normal force doesn’t change (d) 2.8 m/s^2 (e) 28 m/s (f) no (g) 69 N (h)
57 N (i) 40 N (j) 0.33 (k) 0. 09
29.. - zero
2.−kx 0 - (b)f 1 =μkm 1 g cosθ;f 2 =μkm 2 g cosθ(c) Ma (d)TA= (m 1 +m 2 )(a+μcosθ)andTB=m 2 a+μm 2 cosθ(e)
Solve by usingd= 1 / 2 at^2 and substitutinghford - Yes, because it is static and you know the angle andm 1
- Yes,TAand the angle gives youm 1 and the angle andTCgives youm 2 ,m 1 =TAcos 25/g andm 2 =
TCcos 30/g - (a) 3 seconds d. 90 m
32..
33..
34.. - 1.5 N; 2.1 N; 0. 71
Ch 6: Centripetal Forces
1..
2..
3..
4..
- 100 N
- 10 m/s^2
- 25 N towards her
- 25 N towards you
- 14.2 m/s^2
- 1 × 103 N
- friction between the tires and the road
5..0034g
- 2 × 105 m/s^2
- The same as a.
- 56 × 1022 N
- 2 × 10 −^7 N; very small force
8.g= 9 .8 m/s^2 ; you’ll get close to this number but not exactly due to some other small effects
- 2 × 10 −^7 N; very small force
- (a) 4× 1026 N (b) gravity (c) 2× 1041 kg
10..006 m/s^2
1.. 765 - 4880 N