Newton’s 2nd Law: Class Work



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Class Work



  1. A 2000 kg car travels in a straight line on a horizontal road. The relationship between car’s velocity and the time are given by the above graph.




    1. What is the car’s acceleration during first 20 s?

    2. What is the net force applied by the engine during the first 20 s?

    3. What is the car’s acceleration from 20 s to 40 s?

    4. What is the net force applied by the engine during this time?

    5. What is the car’s acceleration from 40 s to 50 s?

    6. What is the net force applied by an engine during this time?


Homework



  1. A 180 kg motorcycle travels in a straight line on a horizontal road. The relationship between motorcycle’s velocity and time are given by the above graph.




    1. What is the acceleration during the first 5 s?

    2. What is the net force during first 5 s?

    3. What is the acceleration from 5 s to 10s?

    4. What is the net force from 5 s to 10 s?

    5. What is the acceleration from 10 s to 15 s?

    6. What is the net force from 10 s to 15 s?

    7. What is the acceleration from 15 s to 25s?

    8. What is the net force from 15 s to 25 s?


Answers

  1. 0.92 N

  2. 4.07 m/s2

  3. -12.25 N

  4. 35,651 N

  5. 0.84 N

  6. 3.96 m/s2

  7. 107.8 kg

  8. -17,000 N

  9. 2361 N

  10. 75 N

  11. 1818 kg

  12. 5.875 m/s2

  13. 162 N

  14. 3.46 s

  15. 558.6 N

  16. 13.6 kg

  17. 75 kg

  18. -80 N

  19. 13.9 m/s

34.75 m

  1. 59.2 kg

  2. 19600 N

  3. 833 N

136 N

No


  1. 1479.6 kg

  2. 44.1 N

  3. 3.47 kg

3.47 kg

13.2 N


  1. 27.6 kg

  2. 588 N

  3. 43.8 N

7.16 N

No


  1. 0.0196 N

  2. 33.7 kg

  3. 8192.8 N

3176.8 N

32-37) see last page



  1. 5 N

  2. 0.6

  3. 12.5 N

  4. 0.077

  5. 32 N

43) 70.56 N

44) 17.1 N

45) 31.9 kg


  1. 0.42

  2. 7.84 N




  1. 153.3 N

  2. 0.39

  3. 64.3 kg

  4. 20 N

  5. 23.52 N

  6. 0.09

  7. 42.5 kg

  8. a) balanced

b) 20 N

c) 0 m/s2



  1. a) balanced

b) 40 N

c) 0 m/s2



  1. a) balanced

b) 60 N

c) 0m/s2



  1. a)

b) 37.5

c) 4.167 m/s2



  1. a) FN

FFr Fapp

mg
b) 300 N



c) 0 m/s2

  1. a) FN

FFr Fapp

mg
b) 500 N

c) 0 m/s2


  1. a) FN

Ffr Fapp

mg
b) 750 N



c) 0 m/s2

  1. a) FN
    Ffr
    mg Fapp
    b) 375 N

c) 4.5 m/s2

  1. a) balanced

b) 5 N

c) 0 m/s2



  1. a) balanced

b) 10 N

c) 0 m/s2

65) a) balanced

b) 16 N

c) 0 m/s2

66) a) balanced

b) 20 N


c) 0 m/s2

  1. a) 14700N

b) 14700N

c) 16500 N

d) 12900 N


  1. 1.63 m/s2

  2. a) 637 N

b) 637 N

c) 793 N


d) 481 N

  1. -1.84 m/s2

  2. a) 548.8 N

b) 228

c) 649.6 N



  1. a) 1372 N

b) 1316 N

c) 1428 N



  1. ­+0.43 m/s2

  2. +0.84 m/s2

  3. a) 0.75 m/s2

b) 350 m

c) net force in

direction of Fapp

d) 18750 N



  1. a) net force in direction of Fapp

b) 205.9 N

c) 1.37 m/s2

d) 31 m/s

e) 22.6 s



  1. a) m1 FN a=
    FT
    m1g

    m2 FN a=


    FT FA
    m2g
    b) a = 3.5 m/s2

c) T = 1.4 N

  1. a) A FN
    FB-A FA
    mAg
    mBg
    FA-B
    mBg
    b) a = 2.08 m/s2

c) T = 33.3 N

  1. a) FT FT
    m1g

    m2g


    b) a = 1.1 m/s2

c) T = 130.5 N

d) x = 4.95 N

e) v = 5.5 m/s


  1. a ) FT FT
    m1g
    m2g
    b) a = 0.96 m/s2

c) T = 39.8 N

d) x = 3 m

e) v = 4.8 m/s


  1. b) FT-FFR=m1a
    c)FT-m2g=-m2a
    d) a = 2.14 m/s2

e) T = 2.3 N

  1. b) FT =m1a
    c)FT-m2g=-m2a
    d) a = 3.33 m/s2

e) T = 2.664 N

  1. b) F = 269.5 N

d) F = 543.9 N

f) F = 583.9 N



  1. a) 1 m/s2

b) 2000 N

c) 0 m/s2

d) 0 N

e) -2 m/s2



f) -4000 N

  1. a) 2 m/s2

b) 360 N

c) 0 m/s2

d) 0 N

e) -1 m/s2



f) -180 N

g) -0.5 m/s2

h)-90 N



Fn

mg
32)

33)


Fsled Fapp

Fn
mg

Fn

mg



Fn
Ffr

mg
34) a)


b)
c)
35)
36)
37)

Dynamics Chapter Problems - v 2.0 ©2009 by Goodman & Zavorotniy



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