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Homework Help: Ramp, block, and elastic potential energy

  1. Apr 18, 2012 #1
    1. The problem statement, all variables and given/known data

    The spring in the figure below is initially compressed by 0.4 m in the position shown. If released from this position, block A travels 0.9 m before coming to a stop. The kinetic coefficient of friction is 0.8. What is the spring constant? (The spring is not fastened to block A)


    2. Relevant equations

    Elastic potential energy=0.5kx^2

    3. The attempt at a solution

    The force of gravity acting on the block=9.81*4=39.24 N

    The force of gravity acting on the block parallel to the ramp=sin(10°)*39.24=6.81 N

    The force of gravity acting on the block perpendicular to the ramp=cos(10°)*39.2=38.64 N, which is also what the normal force is equal to.

    Force of friction=38.64*0.8=30.91 N

    Elastic potential energy=(0.5)(0.4)^2(k)

    The block will gain kinetic energy from the elastic potential energy + the gravitational potential energy, and lose an amount equal to that (since it stops) due to friction, so:

    elastic potential energy+gravitational potential energy=energy lost to friction

    However, the answer I'm given is k=260. I'm not sure if this is a typo, rounding error, or if I'm doing something wrong. Any help would be much appreciated.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Apr 18, 2012 #2
    You're very close. Your mistake is in the gravitational potential energy. You are told that it travels 0.9 m along the surface; so is the block initially 0.9m vertically above the final state, or is it a different amount?
  4. Apr 18, 2012 #3
    I found the gravitational energy by multiplying the amount of gravitational force acting parallel to the ramp (6.81 N) by the distance it acts for:

    ≈6.129 J

    Using the gravitational potential energy equation, I get the same answer:

    ≈6.13 J
  5. Apr 18, 2012 #4
    Try solving it smply.

    Work done on the box = Change in K.E
  6. Apr 18, 2012 #5
    My apologies. I saw no sin(10) in the last line and assumed you had overlooked it. The work looks correct to me otherwise.
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