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Up incline

  1. Aug 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Interesing...

    A spring (k=80n/m ) has an equilibrium length of 1.00m . The spring is compressed to a length of 0.50m and a mass of 2.1kg is placed at its free end on a frictionless slope which makes an angle of 41 degrees with respect to the horizontal. The spring is then released.

    a) If the mass is not attached to the spring, how far up the slope will the mass move before coming to rest? b) If the mass is attached to the spring, how far up the slope will the mass move before coming to rest? c) Now the incline has a coefficient of kinetic friction. If the block, attached to the spring, is observed to stop just as it reaches the spring's equilibrium position, what is the coefficient of friction?





    2. Relevant equations


    I think the equation is: .5mvi^2 + mgyi + .5 kxi^2 = .5 mvf^2 + mgyf + .5 kxf^2
    3. The attempt at a solution

    a) The initial and final velocities are zero. You don't need the final spring configuration because the block leaves the spring.
    so
    mgyi + .5kxi^2=mgyf
    so
    2.1(9.8)(.5sin41) + .5(80)(.5^2) = 2.1(9.8)y?

    do I want y as the answer or x? and if I want x would y = xsin41?

    b) The initial and final velocities are zero.

    c) Use Work done by friction = Change in PE (again no change in KE)
     
  2. jcsd
  3. Aug 17, 2009 #2
    Well your method looks correct for parts a-c. I think that part a is asking for the distance up the ramp so I think figure out the x and y distances and use pythagora's.
     
  4. Aug 17, 2009 #3

    Redbelly98

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    Just to clarify (because I'm not sure it is clear yet), they are asking for the distance along the slope, not the vertical or horizontal distance.

    Also, the question asks to find how far the mass moves. I interpret that to mean the distance between the mass's final position and its initial position. So if you take "0" as the fixed end of the spring, the answer is not simply the mass's final coordinate.
     
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