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Least amount of time for motion

  1. Oct 3, 2015 #1
    1. The problem statement, all variables and given/known data:

    The coefficient of static friction is 0.65 between two blocks. The coefficient of kinetic friction between the lower block and the floor is 0.20. Force Fvec.gif (a pull applied to the top block) causes both blocks to cross a distance of 5.0 m, starting from rest. What is the least amount of time in which this motion can be completed without the top block sliding on the lower block?

    Let:
    topBlock = m1
    lowerBlock = m2

    Given:
    m1 = 4.0 kg
    m2 = 3.0 kg
    μs = 0.65
    μk = 0.2
    Δd = 5.0 m

    2. Relevant equations:
    ∑F = ma
    Δd = (1/2)at^2


    3. The attempt at a solution:

    Starting with Δd = (0.5)at^2

    a = [2Δd]/[t^2]

    Now using ∑F = ma

    ΣF = [m1+m2]*[a] = [m1+m2]*[2Δd]/[t^2]
    ΣF = [2Δd*(m1+m2)]/t^2
    t^2 = [2Δd*(m1+m2)]/[ΣF]

    t^2 = [2*5*(4+3)]/[ΣF]
    t^2 = [70]/[ΣF]

    To find ΣF I said that the force applied must be one that is

    m1gμs > F > (m1+m2)gμk → (4)*(9.8)*(0.65) > F > (4 + 3)*(9.8)*(0.2) → 25.48N > F > 13.72N

    Then I use 25.48N and

    t^2 = 70/25.48
    t = √[70/25.48]
    t= 1.657 s

    And it's wrong :(
     
  2. jcsd
  3. Oct 3, 2015 #2

    andrewkirk

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    Science Advisor
    Homework Helper
    Gold Member

    Your calculation looks correct to me. Maybe somebody else can spot a problem we missed, or maybe the answer in the book is wrong.
     
  4. Oct 3, 2015 #3

    OmCheeto

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    Gold Member
    2016 Award

    I also got the same answer.
     
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