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ksaimouli

  • one year ago

solve

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  1. ksaimouli
    • one year ago
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    |dw:1360985504635:dw|

  2. ksaimouli
    • one year ago
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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?

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

  4. Diwakar
    • one year ago
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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

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

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

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

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

  9. ksaimouli
    • one year ago
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    is this right

  10. ksaimouli
    • one year ago
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    for (a)

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

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

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

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

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

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

  17. Diwakar
    • one year ago
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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

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

  19. ksaimouli
    • one year ago
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    u mean that

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

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

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

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

  24. ksaimouli
    • one year ago
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    u mean that right

  25. ksaimouli
    • one year ago
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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

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

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

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

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

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

  31. ksaimouli
    • one year ago
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    @Diwakar

  32. Diwakar
    • one year ago
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    yes for part b

  33. ksaimouli
    • one year ago
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    okay thx very much

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