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Plz HELP: A block of mass 2.5 kg is placed against a compressed spring

  1. Jul 28, 2013 #1
    1. The problem statement, all variables and given/known data

    A block of mass 2.5 kg is placed against a compressed spring (k = 2900 N/m) at the bottom of an inclined plane. When the spring is released the block is projected up the incline and the spring expands by 14 cm to its normal length.
    Calculate the maximum distance traveled by the block up the incline.

    ** the angle of the incline is 28 degrees.

    a.without friction.

    b. Repeat the calculation with friction, taking µk = 0.22.

    2. Relevant equations

    E=1/2kx^2
    E=1/2mv^2
    E=mgy + 1/2kx^2
    sin(theta)=opp/hyp
    xtotal= x1+x3

    3. The attempt at a solution

    for a) I seperated the question into 3 situations: the compressed block, the block at normal length, and the block at its highest point

    Situation 1: let y=0m here
    E1= 1/2kx^2

    Situation 2: let x=0m here
    E2=1/2mv^2

    Situation 3:
    E3= mgy + 1/2kx^2

    I used Situation 1 and 3...
    E1=E3
    1/2kx1^2 = mgy3 + 1/2kx3^2

    I used trig ratios to find that y3= 0.14sin28

    I solved: x1 = 0.0207 , then added it with x3:
    xtotal= x1+x3 = 0.14m - 0.0207m

    I feel like I'm doing this completely wrong. Can someone explain step-by-step the correct way to solve for this?

    b) I know there has to be non-conservative work done, but I have no clue which situation to use... I'm confused, please help !!

    & THANK YOU
     
    Last edited: Jul 28, 2013
  2. jcsd
  3. Jul 28, 2013 #2

    rock.freak667

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    °
    I think you have over complicated the situation here! Also where did you get the 28° in the sin(28)?


    Let's take each line and see what they are saying:

    So basically you know the mass and the spring constant (m & k).

    So this is telling you what about the extension 'x' of the spring?

    Now if there is no friction, then the all of th elastic energy of the spring is converted to potential energy for the block to reach its maximum height.

    So you will have E1=E2 or mgh = ½kx2
     
  4. Jul 28, 2013 #3
    You need the angle of the incline to calculate the distance. Without it, you can only calculate the max elevation in case a). Speaking of which, how is the max elevation related to the initial state of the spring?
     
  5. Jul 28, 2013 #4
    ** sorry, forgot to include the angle; the angle of the incline is 28 degrees.
     
  6. Jul 28, 2013 #5
    yeaah, i definitely over complicated things. i figured out how to solve for part a).

    for part b), do I calculate the non-conservative force (Wnc= Ffdcos180) from the bottom or the top of the incline? How do you know?

    E1 + Wnc= E2
    or
    E1= E2+ Wnc <-- this makes most sense for me right now ...

    Thank you !.
     
    Last edited: Jul 28, 2013
  7. Jul 28, 2013 #6

    rock.freak667

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    Well your spring is at the bottom of the incline, so your 'd' is measured from the bottom. Your Wnc should be μFfdcos18, as it is friction you are dealing with.
     
  8. Jul 28, 2013 #7

    PhanthomJay

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    Energy from work done by non conservative forces can be transferred into or out of the System. Thus sometimes the work can be positive and sometimes it is negative.
    Always use your first equation, and use the correct signage for the work done by the non conservative force.
     
  9. Jul 28, 2013 #8
    So would the equation for part b) be:

    E1+Wnc= E2
    1/2kx^2 + Ffdcos180 = mgy
    1/2kx^2 - μFnd = mg(dsin28)
    1/2kx^2= d(mgsin28) + μmgcos28 d
    1/2kx^2 = d(mgsin28 + μmgcos28)
    d= 1/2kx^2/ mg(sin28+μcos28)
    d= 1.75 m

    Correct me please; glad to be apart of this forum :)
     
  10. Jul 28, 2013 #9

    haruspex

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    Looks right.
     
  11. Jul 28, 2013 #10
    Thank you to everyone who responded !
     
  12. May 28, 2014 #11
    I realize this is an old thread but I am having trouble understanding part A. Can anyone help?

    if i use: mgh = ½kx2 do i find "h" and then use h to determine the length it travels up the ramp (ie. h/sin28=max dist ??)
     
    Last edited: May 28, 2014
  13. May 29, 2014 #12

    PhanthomJay

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    Yes, that is correct.
     
  14. Aug 23, 2014 #13
    isn't the energy due to 1/2kx2 directed at an angle 28; how can it be equal to mgh? I am wondering if i should use the y component of the 1/2kx2?
     
  15. Aug 23, 2014 #14
    Welcome to Physics forum.
    Energy is not a vector quantity.How can you break it into components?
     
    Last edited: Aug 23, 2014
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