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What faults I have made in the following mathematical soluti

  1. Apr 7, 2016 #1
    I didn't want solution of the following problem. When I was trying to solve the following problem, I found some difficulties that should be solved with the help of this forum.

    We can consider a substance of 0.5kg. It hits a horizontal spring with a velocity of 4 m/s. As a result spring gets contracted and this contraction is 2 meter when the velocity of that substance of 0.5 kg becomes zero. We will find out spring constant.

    Simply I solved in the following way:

    0.5kx2=0.5mv2
    Or, k (22)=0.5 (42)

    Or, k=2 N/m.

    Do you think that this procedure is right?
    If I want to follit problem in an alternative way, I think, we might follow the following way.

    v2=u2-2as

    Or, 0=42-2a.2
    Or, a= 4 m/s²
    We know the value of accelearation, a, increases with an increase of displacement of spring from its equilibrium state. When contraction reaches to its maximum point, the value of acceleration will be maximum. And this 4 m/s² denotes the value of accelaration when spring reaches to its lowest point, i.e when contraction is 2 meter.

    So we can write, when x=2, force that is generated in spring is kx. And it is equal to ma.

    kx=ma
    Or, 2k=(0.5).4
    Or, k=1N/m.

    These two methods two different valuse. What is the wrong? Please help me to get from it.

    I want to mention another problem related to it. That may help you to understand what is in my mind.

    Let, contraction of a spring is 4 cm.
    Mass of a bullet is 10 g.
    Spring constant is 200 N/m.

    When bullet leaves the gun, what will be the velocity of it?

    First I solved it in the following way:

    0.5mv²=0.5kx²
    Or, mv²=kx²
    Or, v=5.6569 m/s².

    But if I want to solve it in the second way, I may write:

    F=kx
    Or, ma=kx
    Or, 0.01a=(200)(0.04)
    Or, a=800 m/s.

    Again we know v²=u²+2as
    Or, v=8 m/s.

    Which is totally different from the first method. In this case I know what the wrong is. I need to use the average value of acceleration. If we use 400m/s² instead of 800 m/s², we will get again v=5.66 m/s.

    I have no problem with the second mathematical problem. I have problem with the first one, and I hope I have already cleared what my problem is.

    Now I want solution from the expert of this forum.

     
  2. jcsd
  3. Apr 7, 2016 #2

    DrClaude

    User Avatar

    Staff: Mentor

    I using a constant acceleration compatible with Hooke's law?
     
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