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Using Laws of Motion

  1. Jan 24, 2008 #1
    Using Laws of Motion!!!

    1. The problem statement, all variables and given/known data
    The acceleration of a brick when lifted by 42N is a. The acceleration of the same brick lifted by 52N is 2a. What is the weight of the brick?

    2. Relevant equations

    3. The attempt at a solution
    I don't really know where to start, here's what I've done so far:
    Fapp=ma, m=F/a



    hopefully that's a good start. thanks
    Last edited: Jan 24, 2008
  2. jcsd
  3. Jan 24, 2008 #2


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    Homework Helper
    Gold Member

    Thats a great start!

    So, you have two equations and two unknowns. A good rule to keep in mind is that if you have the same number of equations and unknowns, then it is possible to find both unknowns. Thus, you should be able to solve for W, and a, if you wish to.

    HINT: You have to use the two equations for W you derived to get rid of a. Can you find a in terms of W? If so, you should be able to plug that in for a in one of those equations.
  4. Jan 24, 2008 #3
    Thx Go1. Glad to know I'm on the right track.
    I was wondering if I could get some clarification on getting rid of a...
    if I can use my two W equations to isolate for a
    I have:

    or in general: a=Fg/W

    I attempted to plug in the first 'a' equation into the second:


    buut, i don't think that's the way to do it since W cancels out...h'mm..
    Last edited: Jan 24, 2008
  5. Jan 24, 2008 #4
    for some reason my edit button isn't working....
    but, to just show what I mean with the W equations:


    I'm sorry, but I think that i just might've understood your question wrong.
  6. Jan 24, 2008 #5


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    Gold Member

    OK I think I see why the weight is canceling out above. The equations we have aren't correct. (I missed the mistake as well.)

    Think of it this way, when a block is lifted by 42N:

    [tex]\Sigma F=ma[/tex]

    But, [tex]\Sigma F= 42N-W[/tex]

    Set up the equations from Newton's second law remembering that we need the NET force on the block as it is lifted up, not just the applied force. See if you can get an answer going about the problem this way.
  7. Jan 24, 2008 #6
    so now to confirm...


    Last edited: Jan 24, 2008
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