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Block moving down an inclined plane, spring problem.

  1. Sep 10, 2008 #1
    1. The problem statement, all variables and given/known data


    Find the amount which the spring is compressed after the mass comes to rest. Assume a kinetic coefficient of friction of 0.11 and an initial velocity along the surface of v = 4m/s.

    Given values :
    mass = 2.2kg
    k = 500 N/m
    delta = 0.036m
    theta = 23 degrees

    The attempt at a solution

    Alright.. here goes

    My approach to this problem is to find the velocity at where the block touches the spring, and make it as the initial velocity.
    Using a free body diagram, I got the following values.

    F = mg = 21.56N
    N = mg cos theta = (21.56)cos23 = 19.846N
    friction force = N.coefficient = (19.846)(0.11) = 2.183N
    force down the plane = mg sin theta = (21.56)sin23 = 8.424N

    Adding up the force in the x direction (parallel to the slope), F = 8.424 + (-2.183) = 6.241N

    since F = ma,
    6.241N = (2.2kg)a
    therefore, a = 2.837m/s^2 [sorry, i'm not familiar with the forums. the unit is metre per seconds squared]
    Now, with the acceleration, the velocity at when it touches the spring can be calculated :

    final velocity squared = initial velocity squared + 2(acceleration)(distance)
    v^2 = u^2 + 2(a)(r)
    v^2 = 4^2 + 2(2.837)(0.036)
    v = squareroot(16 + 0.204)
    v = 4.025m/s

    therefore, the velocity at when it touches the spring is 4.025m/s
    Using the work-energy equation,

    T1 + U = T2

    as when the final velocity is when the block comes to rest, the final velocity is 0, therefore

    (mv^2)/2 + U = 0
    17.821 +U = 0
    therefore U = 17.821J
    Using the value, the compression length can be found:
    17.821 = 0.5(k)(final compression length squared - initial compression length squared)
    17.821 = 0.5(500)(x^2 - 0)
    17.821 = 250(x^2)
    0.071283 = x^2
    x = squareroot(0.071283)
    x = 0.267m

    that was my answer, but it's wrong. Much help would be appreciated on where i went wrong and how to solve the problem. Thanks.
  2. jcsd
  3. Sep 10, 2008 #2
    What is T2 and why is it zero? Where are you considering the potential energy of the block in this energy balance?

    If I was to go about this problem, I would do it in two steps. First would be to find the velocity of the block at the spring. a = (Fg - Ff)/m

    Then I would do an energy balance of the spring and block as a system. KEblock + PEblock = PEspring

    Which you would solve for the displacement in the PEspring term.
  4. Sep 10, 2008 #3
    Thanks for the tip, but i don't quite get your formula for acceleration. Can you explain more?

    For the potential energy, yeah, i forgot about that. After revising my calculations, i still got the wrong answer.

    Kinetic energy of block = 0.5mv^2 -0.5mu^2 (where v = final velocity and u = initial velocity)
    therefore kinetic energy = -17.821
    potential energy of block = -mg(0.036sin23) = -0.303

    potential energy of spring = -0.5kx^2

    solving the equation I ended up with 0.269m, which is still wrong :(
  5. Sep 10, 2008 #4
    The solution is 0.2798m.

    How is it possible? working backwards I found the acceleration to be around 21m/s^2. Isn't the acceleration suppose to be less than gravity? Can someone help me point out where i went wrong please? (please refer to my calculations above for my steps on finding the acceleration). Much help would be appreciated!
  6. Sep 11, 2008 #5


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    Homework Helper

    Hi Dealucis,

    I don't believe this is correct. You're writing the energy equation for the motion while the spring is being compressed; but the distance 0.036 refers to the part of the motion that came before that. What do you think this term should be?

    I'm not seeing where you are getting the acceleration to be 21m/s^2 if you use that answer. Can you give more details?
  7. Aug 22, 2009 #6
    You are missing the loss of energy in this part.

    So, the equation should be U-T1=Wf (where U = potential energy of spring, T1=total initial energy of object, Wf= work done by friction)

    U is equal to T2 ( T2 = total final energy of object)
    It is because when the spring is compress by object, so the spring will gain the potential energy from the total final energy of object.

    0.5*2.2*(4.025^2) + 2.2*9.8*(x*sin(23))=2.185*x + 0.5*500*x^2

    From above equation u will get the value of x which is the amount which the spring is compressed.

    x= 0.2798m
    Here is the answer.
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