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Work Energy Problem

  1. Nov 9, 2015 #1
    1. The problem statement, all variables and given/known data
    The 2 kg piece of putty is dropped 2m onto the 18 kg block initially at rest on the two springs, each with a stiffness k =1.2kN/m . Calculate the additional deflection of the springs due to the impact of the putty, which adheres to the block upon contact.

    2. Relevant equations
    W = ΔKe + ΔEp +ΔEs

    3. The attempt at a solution
    I started by saying that there is no external force so W = 0.
    Next I found the speed of the block just as the putty hits it.
    v1 = putty velocity = v^2 = 2*9.8*2
    v = 6.26m/s

    P1 = P2
    m1v1 = (m1+m2)v2
    2*6.26 = (2+18)v2
    v2 = 0.63m/s

    ΔKe = 1/2m(Δv^2)
    = 1/2 * 20 * (0 - 0.63^2) (max defl at v = 0)
    =-3.92J

    ΔEp = mgΔh
    =20*9.8*δ
    =196δ J

    ΔEs = 1/2k(x^2 - x^2)*2
    = 1200((0.0735+δ)^2-(0.0735)^)
    =1200(0.147δ +δ^2)
    =176.4δ+1200δ^2

    0 = 1200δ^2+372.4δ-3.92
    Using quadratic formula
    δ=0.204m

    Just want to know if this is right, as I don't think it is. Also would be great if you could tell me where I went wrong. Thanks
     
    Last edited: Nov 9, 2015
  2. jcsd
  3. Nov 10, 2015 #2

    Simon Bridge

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    To troubleshoot: at each stage, write a short statement describing the energy change, or other process, and keep the variables... do the algebra 1st then plug the numbers in.
    i.e. step 1, putty ##m_1## falls distance ##h## exchanging gpe for ke... so ##v_1^2 = 2gh/m_1## ... gaining momentum ##p = \sqrt{2ghm_1}## ... which is conserved... kinetic energy is ##p^2/2m## ...

    Take care to be consistent with variable names... ie k is already used for a single spring.
    This should tell you if and ehere you have made a mistake.
     
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