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Motion with elastic potential energy and other forces

  1. May 30, 2016 #1
    1. The problem statement, all variables and given/known data
    A 15.0 kg stone slides down a snow-covered hill leaving point A (at the top of the hill) with a speed of 10.0 m/s. There is no friction on he hill between point A and point B (at the foot of the hill). There is friction on the level ground at the bottom of the hill between point B and a wall some distance away. After entering the rough horizontal region, the stone travels 100 m and the runs into a very long light spring with force constant 2.00 N/m.
    The coefficients of kinetic and static friction are 0.20 and 0.80, respectively.

    a) What is the speed of the stone when it reaches point B ?

    b) How far will the stone compress the spring ?

    c) Will the stone move again after it has been stopped by the spring ?

    2. Relevant equations
    E = K + U
    K1 + Ugr = K2 + Ugr2
    K1 + U1 + W (friction) = K2 + U2
    K1 + Uel1 + W(friction) = K2 +Uel2

    Where K1, K2 are initial and final Kinetic Energy,
    U1, U2 are initial and final Potential Energy
    Ugr1, Ugr2 initial and final Gravitational Potential Energy
    Uel1, Uel2 initial and final Elastic potential Energy
    W(friction) = work done by Friction (= (f) x (distance))
    Uel = 1/2 k x(sqrd)

    3. The attempt at a solution
    Now I solved part a) with little difficulty noting that since this area has no friction then the stone's gravitational potential energy is entirely converted into kinetic energy by the time it reaches ground-level , so
    mass x gravity x height = 1/2 m v(sqrd) (we also have a KE value for initial speed) ... the speed = 22.2 m/s .. which agrees with the book.

    However the stumbling block is part b) .. no matter what I do I cannot get the correct answer for this part of the problem.

    I attempted it as follows ..
    I took it in 2 parts .. first from point B to the spring and secondly from touching the spring to full compression.

    Initially I calculated speed on reaching spring using ..
    v(sqrd) = v(initial)(sqrd) + (2a) x (distance) (a calculated from friction f=ma).

    Next I noted that since we have a speed value there must be an initial KE value too. Also there is now no gravitational PE value (no height) this is zero.
    the other forces at work here comprise of kinetic friction alone.

    So using the general energy equations here I got
    K(1) + U(1) + W(friction) = K(2) + U(2)

    Where K(1), K(2) are initial and final kinetic energies of stone
    U(1), U(2) are initial and final potential energies of stone
    W(friction) is work done by friction (f x distance (100 m))
    U(1) and U(2) are zero ...
    from this we get a value for final KE which I use as initial KE for next phase which is encountering the spring.

    Now at the spring (100 m into the stone's travel) using the energy equation for elastic potential energy ..

    K(1) + Uel(1) + (-W(friction)) = K(2) +Uel(2)

    where K(1) here is just the K(2) value from previous calculations
    K(2) = 0 (stone comes to rest)
    Uel(1) = 0 (spring uncompressed)
    Uel(2) = 1/2 k x(sqrd)

    This gives .. 1/2 m v(sqrd) - [(0.2) mgx] = 1/2 (2.0) x(sqrd)
    = ((0.5)(15.0)(10.4)(sqrd) - 29.4x = x(sqrd)
    = 811.2 - 29.4x = x(sqrd)

    NOTE - The friction term is a term in x as I am calculating the work done by friction over the unknown spring-compression distance.

    The way I attempted to solve this is by rearranging the terms to give a quadratic equation .. x(sqrd) +29.4x - 811.2 = 0
    then solve using the rule ... ax(sqrd) + bx+ c use
    b+/- sqrt((b(sqrd) - 4ac)/2a)
    However I get nonsense solutions when I do this.

    I am at a loss as to where to go from here .. can anyone help please.

    JACKTHEHAT
     
  2. jcsd
  3. May 30, 2016 #2

    TSny

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    I believe the 10.4 should be 10.04.

    Solving this quadratic equation should not give you nonsense, although it does imply that the spring is very long (as stated in the problem).

    Note, you can save yourself some work by setting up work and energy between point B and the point of final compression of the spring. (Or you can set it up between the initial point on top of the hill and the final point.) There is no need to find the KE just as the stone makes contact with the spring.
     
  4. May 31, 2016 #3
    Hi Tsyn,
    Thank you for your input on this one. However I am still stumped on this one.
    You say that I should set up work and energy system between the point B and the final point. I assume by final point you mean at the maximum compression of the spring (ie. 100.0 m + compression distance). I cannot see how I can do that as I do not know how much distance the spring will be compressed by .. that is what I am supposed to find. I can estimate how much work friction does to the stone as I know the coefficient of kinetic friction and I know a distance (100.0 M) but this distance is from point B to the spring not to the final point of compression. So then how can I calculate the full work friction does as I do not have the entire distance traveled to full compression of the spring ?
    When I use the vale of friction x 100.0 m in my equations my final answer as to how much the spring is compressed is .. 26.8 m
    The correct answer from the book is ... 16.4 m.
    So the difference between the two values above has to be the added negative work of the friction force over the compression distance.
    My trouble is I do not know how I can calculate this as I have only an incomplete distance (from point B to the spring before compression).
    Do you see my problem ?
     
  5. May 31, 2016 #4

    TSny

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    Express the total distance traveled between B and final max compression in terms of x. Then you can express the work done by friction between these two points in terms of x. When you set up your work-energy equation, you should get the same quadratic equation as you get doing it your first way.
     
  6. May 31, 2016 #5

    TSny

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    If you change 10.4 to the correct value of 10.04, then the value of 811.2 will change. The solution of the corrected quadratic equation should give you the correct answer.
     
  7. May 31, 2016 #6
    Thank you for all your help Tsny much appreciated.
    Regards,
    JacktheHat
     
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