1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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)

    Zqbu5.png

    2. Relevant equations

    Elastic potential energy=0.5kx^2
    Work=Fd

    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
    (0.5)(0.4)^2(k)+(6.81)(0.9)=(30.91)(0.9)
    k≈271

    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:

    ≈(6.81)(0.9)
    ≈6.129 J

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

    =mgh
    =(4)(9.81)(sin(10)*0.9)
    ≈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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Ramp, block, and elastic potential energy
  1. Elastic energy and ramps (Replies: 16)

Loading...