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ksaimouli

solve

  • one year ago
  • one year ago

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  1. ksaimouli
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    |dw:1360985504635:dw|

    • one year ago
  2. ksaimouli
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    a spring with k 170 N/m is at the top of a frictionless incline of angle 37.0°.The lower end of the incline is distance D 1.00 m from the end of the spring, which is at its relaxed length. A 2.00 kg canister is pushed against the spring until the spring is compressed 0.200 m and released from rest. (a) What is the speed of the canister at the instant the spring returns to its relaxed length (which is when the canister loses contact with the spring)? (b) What is the speed of the canister when it reaches the lower end of the incline?

    • one year ago
  3. ksaimouli
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    what i did is |dw:1360985802827:dw|

    • one year ago
  4. Diwakar
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    use energy conservation in both cases. when compressed , E= kx^2/2 + mgh x=0.2m When it is at bottom h = (D+x)sin37

    • one year ago
  5. ksaimouli
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    for h the total distance id 1m+.2m (compressed distance )

    • one year ago
  6. ksaimouli
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    sin37(1.2)=h

    • one year ago
  7. Diwakar
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    yes you are proceeding correctly

    • one year ago
  8. ksaimouli
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    if i solve it i got v=4.20m/s

    • one year ago
  9. ksaimouli
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    is this right

    • one year ago
  10. ksaimouli
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    for (a)

    • one year ago
  11. Diwakar
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    no this will be the answer to part b i.e. the canister has reached the ground

    • one year ago
  12. ksaimouli
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    (b) i got |dw:1360986031295:dw|

    • one year ago
  13. ksaimouli
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    @Diwakar how can that be part b

    • one year ago
  14. Diwakar
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    you have calculated h corresponding to the lower end of the incline

    • one year ago
  15. ksaimouli
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    it doest not matter right because i got height so the box is top so height works

    • one year ago
  16. Diwakar
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    |dw:1360986099560:dw|

    • one year ago
  17. Diwakar
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    oh sorry the height 1.2sin37 is from bottom to the cannister and not to the top of the incline as i have shown in the figure

    • one year ago
  18. ksaimouli
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    |dw:1360986365438:dw|

    • one year ago
  19. ksaimouli
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    u mean that

    • one year ago
  20. Diwakar
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    yes the black line shows the heigth 1.2 sin37

    • one year ago
  21. ksaimouli
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    shouldnt that be 1sin 37 because the spring not compressed

    • one year ago
  22. Diwakar
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    for part a energy conservation will be |dw:1360986436055:dw|

    • one year ago
  23. ksaimouli
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    |dw:1360986652778:dw|

    • one year ago
  24. ksaimouli
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    u mean that right

    • one year ago
  25. ksaimouli
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    okay for part b u need to use h=1.2sin37 and should we consider the part (a) speed as initial speed

    • one year ago
  26. ksaimouli
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    |dw:1360986763295:dw|

    • one year ago
  27. Diwakar
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    you may do that here h = 1*sin37

    • one year ago
  28. Diwakar
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    since the body has already covered 0.2m in part a as it loses contact with the spring

    • one year ago
  29. ksaimouli
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    for part b u mean

    • one year ago
  30. ksaimouli
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    okay so h=1 sin 37 and i can use above equation right with initial speed from part (a)

    • one year ago
  31. ksaimouli
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    @Diwakar

    • one year ago
  32. Diwakar
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    yes for part b

    • one year ago
  33. ksaimouli
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    okay thx very much

    • one year ago
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