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Two blocks of power (2)

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data

    the force is given as F
    the distance pushed is given as D


    2. Relevant equations

    W= F*d
    F=ma
    KE = 1/2 mv^2


    3. The attempt at a solution

    the blocks have the same kinetic energy
     

    Attached Files:

  2. jcsd
  3. Oct 23, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Correct. Since the same work (F*d) is performed, the resulting KE is the same.
     
  4. Oct 23, 2008 #3
    ok for part b it is asking how fast the lighter block is moving when compared to the big block.

    im thinking

    the F*D is the same

    and KE is .5mv^2

    and so

    .5(4m)(v_L)^2 = .5m(v_S)^2

    the m and .5 cancel out

    so i have

    4(V_L)^2=(V_S)^2

    SQRT BOTH SIDE

    and then

    V_S = 2V_L

    so therefore

    the smaller box is moving twice as fast as the big box?

    right?
     
  5. Oct 23, 2008 #4

    Doc Al

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    Staff: Mentor

    Perfect!
     
  6. Oct 23, 2008 #5
    ok

    now both the blocks have the same speed
    and now have the same V

    ok...

    and its asking for distance
    so i have to find out how far each box is pushed

    and
    work of larger box =

    W_L= F*D_L
    W_S=F*D_S

    KE_L = .5(4M)(v)^2
    KE_S = .5(M)(v)^2

    and since W=chainge in KE (KE_initial is 0)

    F*D_L = .5(4M)(v)^2
    (v)^2 = (F*D_L) / .5(4M)

    F*D_S = .5(M)(v)^2
    (v)^2 = (F*D_S)/.5(M)

    (F*D_S)/.5(M) = (F*D_L) / .5(4M)

    .5 M and F cancel out

    D_S = (D_L) / 4
    D_L = 4D_S

    the distance the larger box has to be pushed is 4 times larger than that of the smaller box?
     

    Attached Files:

  7. Oct 23, 2008 #6

    Doc Al

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    Staff: Mentor

    Exactly!
     
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