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Very easy question: why can't I use all 3 of these constant-acceleration formulas?

  1. Sep 28, 2012 #1
    1. The problem statement, all variables and given/known data
    A dockworker applies a constant horizontal force of 80 N to a block of ice on a smooth horizontal floor. The fritional force is negligible. The block starts from rest and moves 11 meter in 5 seconds.

    What is the mass of the block of ice?


    2. Relevant equations
    Newton's second law:
    [itex]\Sigma F=ma[/itex]
    [itex]v^2_x=v^2_{0x}+2a_x(x-x_0)[/itex]
    [itex]v_x=v_{0x}+a_xt[/itex]

    3. The attempt at a solution
    [itex]\Sigma F=80N[/itex]
    [itex]v_x=\frac{11m}{5s}=2.2m/s[/itex]
    [itex]v_{0x}=0m/s[/itex]
    [itex]x=x_0+v_{0x}t+1/2a_xt^2[/itex]

    Now in order to use Newton's second law, I need to find the acceleration.
    My question is, why can't I use all 3 of the formulas?

    When I use the first formula:
    [itex](2.2m/s)^2=0+2a_x(11m)[/itex]
    Which gives: [itex]a_x=\frac{(2.2m/s)^2}{2\cdot 11m}=0.22m/s^2[/itex]

    But if I use the 2nd formula:
    [itex]2.2m/s=0+a_x\cdot 5s[/itex]
    Which gives: [itex]a_x=\frac{2.2m/s}{5s}=0.44m/s^2[/itex]

    And if I use the 3rd formula:
    [itex]11m=0t+1/2a_x\cdot (5s)^2[/itex]
    Which gives: [itex]a_x=\frac{11m}{1/2\cdot (5s)^2}=0.88m/s^2[/itex]

    What's the difference between the three formulas? Why do I get different accelerations?
     
    Last edited: Sep 28, 2012
  2. jcsd
  3. Sep 28, 2012 #2
    Re: Very easy question: why can't I use both of these constant-acceleration formulas?

    Your [itex]V_x[/itex] is average velocity and not final velocity.
     
  4. Sep 28, 2012 #3
    Re: Very easy question: why can't I use both of these constant-acceleration formulas?

    I've updated my 1st post with another equation. According to my book, it says the 3rd equation is the correct with the acceleration being 0.88 m/s^2

    But that is the average velocity as well?
     
  5. Sep 28, 2012 #4

    SammyS

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    Re: Very easy question: why can't I use both of these constant-acceleration formulas?

    [itex]v_x[/itex] is the final velocity of the block,

    whereas [itex]22 \text{m/s}[/itex] is the average velocity of the block.

    For constant acceleration, the average velocity is given by: [itex]\displaystyle v_\text{Average}=\frac{v_x+{v_0}_x}{2}\ .[/itex]
     
  6. Sep 28, 2012 #5
    Re: Very easy question: why can't I use both of these constant-acceleration formulas?

    The third equation is [itex]S = V_{0x}t + \frac{1}{2}at^2[/itex]
    where S = distance, V0x = initial velocity a = constant acc. t = time, no final velocity needed.

    Acceleration is constant therefore, velocity is changing throughout, the actual final velocity is not given in the question and can be calculated using v = u + at to be 4.4m/s
     
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