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Kinetic Energy vs speed

  1. Jan 1, 2010 #1
    1. The problem statement, all variables and given/known data
    If the Kinetic energy of a given mass is to be doubled, its speed must be multiplied by?

    2. Relevant equations

    KE= 1/2mv^2

    3. The attempt at a solution
    I believe the answer is 4, but am not really sure. I figured that if the Kinetic energy should be doubled, then since the v(velocity) on the other hand is squared and the proportionality also seems to be direct then the answer is 2^2=4. Am I right?
  2. jcsd
  3. Jan 1, 2010 #2
    No, that is not right. For problems like this it is worth picking some numbers to substitute in for variables. As a hint, you aren't going to get a whole number.
  4. Jan 2, 2010 #3
    KE=.5*m*V1^2 therefore V1=Sqrt(2*KE/m)

    2*KE=.5^V2^2 therefore V2=sqrt(4*KE/m)

    What must you multiply V1 by to get V2?


    Z=V1/V2= 2/sqrt(2)
  5. Jan 2, 2010 #4
    u can think about it this way ... u have your KE(1) = 1/2mv^2 , so in order to have 2KE(1) which is (2)1/mv^2 .. what u should have with the velocity so that u get this factor of (2) ..
  6. Jan 2, 2010 #5
    I think u have made some mistake here..
  7. Jan 2, 2010 #6
    still confused... is the answer 1/2?, which is it and how can arrive at the right answer.
  8. Jan 2, 2010 #7


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    Homework Helper

    Consider this: you have two kinetic energies, KEi (initial) and KEf (final). These kinetic energies are computed with the same mass, but with different velocities, vi and vf. You're given that KEf is twice KEi, and you're trying to figure out what the relationship between vi and vf is.

    What you've been doing so far is kind of like a guess-and-check method: you start with
    [tex]KE_i = \frac{1}{2}mv_i^2[/tex]
    [tex]KE_f = \frac{1}{2}mv_f^2[/tex]
    and try plugging in first [itex]v_f = 4v_i[/itex] (your initial guess)
    [tex]KE_f = \frac{1}{2}m(4 v_i)^2 = 16*\frac{1}{2}m v_i^2 = 16KE_i[/tex]
    ...nope, that's not it. Now what if you try 1/2?
    [tex]KE_f = \frac{1}{2}m\biggl(\frac{1}{2} v_i\biggr)^2 = \frac{1}{4}*\frac{1}{2}m v_i^2 = \frac{1}{4}KE_i[/tex]
    ...that's not it either.

    Try the algebraic approach. Start with the condition that you need to be true, that the final KE is twice the initial KE
    [tex]KE_f = 2 KE_i[/tex]
    and substitute in the expressions for kinetic energy
    [tex]KE_i = \frac{1}{2}mv_i^2[/tex]
    [tex]KE_f = \frac{1}{2}mv_f^2[/tex]
    Then cancel out the common factors on both sides of the equation and see what you're left with. Finally, what mathematical step should you take to get it into the form
    [tex]v_i = \bigl(\cdots\bigr)v_f[/tex]
    and what is the coefficient in the parentheses? That will be the answer you're looking for.
  9. Jan 2, 2010 #8
    Re: Kinetic Energy ,,

    Thanks a lot for the hint... really appreciate it. so my answer turns out 2^1/2
    I figured it logically: since 'v' is squared, what number when squared gives 2.
    In other words 1.41
  10. Jan 2, 2010 #9
    2/sqrt(2) is 1.41 as described above
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