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Please help with hw problem finding acceleration

  1. Oct 27, 2004 #1
    hw Problem:
    A block of mass 1 kg rests on a tabletop with coeficient of kinetic frictionequal to 0.3. The block is conected by a string which passes over a frictionless pulley to a second block of mass 2kg which hangs vertically from the string. The acceleration of the two block is.....?

    mass(1)=1kg
    mass(2)=2kg
    uk=0.3(coeficient)
    Fr(friction)=uk*Fn(normal force)

    My solution:
    (x)F=ma

    box 1
    component (x)=Ft(force tension)-Ffr(friction)=ma
    component (y)=Fn(normal force)-mg(weight)=m(a=0)
    Fn=mg
    Fn=1kg*9.8=9.8N

    component (x)=Ft-uk*Fn=ma
    component (x)=Ft=ma-uk*Fn
    Ft=ma-0.3*9.8=ma-2.94
    Ft1=ma-2.94N

    Box 2
    component (y)=mg-Ft=ma
    mg-ma=Ft

    Now: Ft1=Ft2
    m(1)a-2.94=m(2)g-m(2)a
    a=(m(2)g+2.94)/m1+m2

    a=(2*(9.8)+2.94)/2+1=7.51m/s^2

    IS THIS SOLUTION RIGHT? If not please indicate where i made a mistake
     
  2. jcsd
  3. Oct 27, 2004 #2

    Ba

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    On box 1 you made a mistake, Ft-uk*Fn=ma then you went to Ft=ma-uk*Fn. It should be Ft=ma+uk*Fn.
     
  4. Oct 27, 2004 #3
    The net force is F=2*9.8+0.3*9.8=22.54 N. The net mass is m=2+1=3kg. So the acceleration is a=F/m = 22.54/3=7.5 m/s^2. Thus your answer is correct
     
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