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Can Kinetic Energy equal 1/2W

  1. Jan 22, 2012 #1
    Ok, so I had to find a way to find Kinetic Energy without finding the velocity, so I decided to move the equation, Ek=.5mv^2. Now then Ek.5m(x/t)^2 where x is distance and t is time. Work on the other hand is W=Fx, and F=ma, W=ma*x and a=(x/t^2), hence, W=m* (xx/t^2) or W=m*(x^2/t^2) or W=m*(x/t)^2. which is also the second part of the Ek equation, therefore, Kinetic Energy can be found by finding half of the work, could this work, is this actually possible or did I do something completely wrong?
     
  2. jcsd
  3. Jan 22, 2012 #2
    The problem is that you're using equations for average values, you need to be doing the actual calculus:
    [tex]W=\int F \hspace{0.05in} dx = \int ma \hspace{0.05in} dx [/tex]
    If acceleration (and mass) is constant...
    [tex] W = ma \int dx = ma*x [/tex]
    So this part is like you said. But, for constant acceleration x does not equal [itex](x/t^2)[/itex] (because the velocity is not necessarily zero)
    [tex]x = v_0 t + \frac{1}{2} a t^2 \rightarrow a = \frac{2}{t^2}(x-v_0 t) [/tex]
    Thus, [tex]W =m \frac{2x}{t^2}(x-v_0 t)[/tex]

    Your expression ends up being correct, if both the acceleration is constant, and the velocity is zero over the entire path (i.e. both the acceleration and velocity are zero), and thus the work and kinetic energy are both zero.
     
  4. Jan 22, 2012 #3

    Pengwuino

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    Gold Member

    Yes, this is not a valid method of manipulating equations. Things like Work are defined like [itex]W = F \Delta x[/itex], a change in a distance, given by [itex]x - x_0[/itex]. This is the same for everything else. Only when the initial position/velocity/whatever is 0 is this even remotely capable of working and it would only work numerically, it's not proper algebraic manipulation.

    The REAL problem in what you have is your definition for acceleration. In a slightly more proper sense (yet still wrong), the acceleration, as you wrote it, would be [itex]a = {{\Delta x}\over{\Delta t^2}}[/itex]. However, this is also [itex]a = {{v_{average}}\over{\Delta t}}[/itex] which, and you can easily verify, is not at all correct.

    Plus, think about this for a second. If a particle were to accelerate from rest to a certain velocity, the work done on the object must EQUAL the kinetic energy, not be 1/2 of work.
     
  5. Jan 22, 2012 #4
    Thank you both, I knew I was wrong, just really wanted an opinion from others since it really seemed like I was just manipulating random equations. Thanks!
     
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