1. Limited time only! Sign up for a free 30min personal 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!

Relating to potential energy of a spring and mass on inclined plane

  1. Oct 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Here is an illustration to ease understanding:
    Angle inclined plane: θ = 20.0°
    spring of force constant k = 525 N/m
    block: m = 2.71 kg
    distance of block from spring: d = 0.330 m
    speed of block: v = 0.750 m/s

    By what distance is the spring compressed when the block momentarily comes to rest? (I'm going to call it 'x' in this problem.)

    The answer should be x = 0.143 m

    2. Relevant equations
    Ok, I'm not quite sure how many of these are relevant or if I'm missing some, but I'll just give some that I know.

    PEs = (1/2)kx2 (x = distance spring is compressed)
    PEg = mgh
    KE = (1/2)mv2
    F = ma
    U + KE = 0 (U = any potential energy? All potential energies? I'm not quite positive what is going on here.)

    3. The attempt at a solution
    Okay. I'm not quite sure how to go about this, so I'll just start doing what I seem to know.

    Let's start with force. It doesn't seem inherently relevant to the problem, but why not?

    Okay, so force in the x direction seems to be: mgsinθ = ma. So we get a = gsinθ = (9.8)(sin 20°) = 3.35198 m/s2.

    Force in the y direction seems to be: N - mgsinθ = 0. So N = mgsinθ = (2.71)(9.8)(sin 20°) = 9.08337 N.


    Ok. That was probably not relevant. Let's do work now.

    Using the numbers above, I get KE = (1/2)(2.71)(.750)2 = 0.76219 J.

    PEg = (2.71)(9.8)h. Okay, I'm not quite sure if h should just be d or if I have to factor in the slope. I'm thinking that slope probably does have to be included, so h is 0.330(sin 20°), so PEg is 2.99751 J.

    So it seems like PEg + PEs + KE = 0. Then it must be that (1/2)kx2 + mgh + (1/2)mv2 = 0... However, solving for x gives me the square root of a negative number, I must be interpreting U + KE = 0 wrong. Where do I go from here? Thanks, I know this is long.
  2. jcsd
  3. Oct 4, 2013 #2
    Why do you think that the energy is zero?
    You can use conservation of energy to solve the problem.
    Equate the total energy in the initial position with the total energy in the final position.
    Total energy here includes kinetic, potential gravitational and potential elastic.
    Some terms may be zero, like elastic energy in the initial position or kinetic energy in the final position.
  4. Oct 4, 2013 #3
    Well, I'm not entirely sure what is meant by the equation U + KE = 0 or if it's relevant...But I do know that a decrease in potential energy would mean an increase in kinetic energy and vice versa.

    So are you saying that PEg = PEs (KE is zero at the initial and final positions)? I'm afraid I don't understand.

    *edit* Hold on a minute I did something weird...

    Okay, so if PEg = PEs, then I get (1/2)kx2 = mgh... and x = √(2mgh/k) = 0.107 m, which is incorrect. If I assume h shouldn't have sin in it, then I get x = 0.183 m. Both of these answers are still wrong. Shouldn't I have to fit velocity into the problem somehow?
    Last edited: Oct 4, 2013
  5. Oct 5, 2013 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    h is the total vertical movement of the block. Don't forget that this will include a contribution from x.
  6. Oct 5, 2013 #5
    No, the initial KE is not zero. It moves with some speed, isn't it?
    And h is measured along the vertical direction.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted