http://www.ck12.org Chapter 26. Answers to Selected Problems
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
30.. - zero
2.−kx 0 - b.f 1 =μkm 1 g cosθ;f 2 =μkm 2 g cosθc. Ma d.TA= (m 1 +m 2 )(a+μcosθ)andTB=m 2 a+μm 2 cosθe. 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
33..
34..
35.. - 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 - 4× 1026 N
- gravity
- 2× 1041 kg
9..006 m/s^2
1.. 765
- 2 × 10 −^7 N; very small force