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Conservation of Energy

  1. Dec 13, 2013 #1

    patrickmoloney

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    1. The problem statement, all variables and given/known data

    A block of mass 5kg is released from rest and slides down a distance D down a smooth plane inclined at 30◦ to the horizontal. It then strikes a spring compressing it 10cm before it begins to move up the plane. If the spring constant is 980 N/m and g=9.8m/s², find D. Solve Using the Principle of Conservation of Energy.



    2. Relevant equations

    mgh=1/2kx²



    3. The attempt at a solution

    I drew a diaghram of the problem and found the resultant forces on the block.

    Kinetic energy of the block= potential at the bottom of the spring.

    mgh=1/2kx²

    m=5 Kg, g=9.8 m/s² , h= Dsin30 (pythagoras theorem) , k= 980 N/m , x= 0.1m

    subbing into the formula

    (5)(9.8)(Dsin30)=1/2(980)(0.1)
    D=0.2 m

    is this the correct solution? it's an assignment so I'd like the method to be correct. Thanks in advance.
     
  2. jcsd
  3. Dec 13, 2013 #2
    You are on the right track, but. You have neglected that the block gains additional energy as it compresses the spring, because it goes lower.
     
  4. Dec 13, 2013 #3

    patrickmoloney

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    how do i show this in mathematics and is D right or wrong due to this additional energy?
     
  5. Dec 13, 2013 #4
    Your basic equation ## mgh = \dfrac 1 2 k x^2 ## is correct. ## h ## should be the height from the top position of the block to the very lowest position, where the spring is fully compressed.
     
  6. Dec 13, 2013 #5

    patrickmoloney

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    the height from the top position of the block to the very lowest position is Dsin30, no? Since d=h/sin(theta)
     
  7. Dec 13, 2013 #6
    No, because the very lowest position is where the spring is fully compressed.
     
  8. Dec 13, 2013 #7

    patrickmoloney

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    is h= (D+0.1)sin30?
     
  9. Dec 13, 2013 #8

    patrickmoloney

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    then if h is indeed =(D+0.1)sin30 , D= 0.15 m
     
  10. Dec 13, 2013 #9
    Correct.
     
  11. Dec 13, 2013 #10

    patrickmoloney

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    Apparently the answer was 0.2?? Where did I go wrong?
     
  12. Dec 13, 2013 #11
    You found previously D = 0.2 m. But now you know that instead of D, you should have used (D + 0.1 m), so D = 0.1 m.

    I do not know why you think the answer should be 0.2 m.
     
  13. Dec 14, 2013 #12

    patrickmoloney

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    i did use h=(D+0.1)sin30 as you said it was correct and got 0.15 m to be the answer. I did not get 0.1 m.

    I thought 0.2 m due to a colleague in the class got this answer and he is better than me at applied maths and he did it a different way to me.
     
  14. Dec 14, 2013 #13
    49(D + 0.1)sin30 = (1/2)980*.1^2
    49Dsin30 + 2.45 = 4.9
    49Dsin30 = 2.45
    49D = 4.9
    D = .1 m

    As Voko said, you find 0.2 when you don't account for the compression of the spring which contributes to the total distance. Your colleague is wrong =P.
     
  15. Dec 14, 2013 #14

    patrickmoloney

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    mgh=1/2kx²

    (5)(9.8)(D+0.1)sin30 = (1/2)(980)(0.1)²
    Dsin30 + 0.1sin30 = (1/2)(980)(0.1)²/(5)(9.8)
    Dsin30 = (1/2)(980)(0.1)²/(5)(9.8) - 0.1sin30
    ..oh I see. I didn't divide - 0.1sin30 by sin30 :(

    schoolboy error. Guys thanks for the help.
     
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