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Homework Help: Double KE, what happens to velocity

  1. Oct 21, 2004 #1
    if you double the KE, by what factor does the speed increase? and vice versa

    i am a little lost
     
  2. jcsd
  3. Oct 21, 2004 #2

    Pyrrhus

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    The speed will increase by [itex] \sqrt(2) [/itex]

    Use the equation

    [tex] K = \frac{1}{2}mv^2 [/tex]

    If the speed increases twice

    [tex] K = \frac{1}{2}m(2v)^2 [/tex]

    [tex] K = 4 (\frac{1}{2}mv^2) [/tex]

    [tex] K = 4K [/tex]

    Kinetic energy will increase four times.

    Now if the kinetic energy increases four times

    [tex] v = \sqrt{\frac{2K}{m}} [/tex]

    [tex] v =\sqrt{\frac{(4)2K}{m}} [/tex]

    [tex] v = \sqrt{4} \sqrt{\frac{2K}{m}} [/tex]

    [tex] v = 2 \sqrt{\frac{2K}{m}} [/tex]

    [tex] v = 2v [/tex]

    The speed will increase two times.
     
    Last edited: Oct 22, 2004
  4. Oct 22, 2004 #3

    robphy

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    To clarify Cyclovenom's reply (i.e. to avoid statements like [tex] K = 4K [/tex], which implies that [tex]K=0[/tex]), try this.

    The relation between kinetic energy and speed is
    [tex]K=\frac{1}{2}mv^2[/tex].

    To double the speed, think [tex] v_\text{new}=2 v_\text{old} [/tex].
    Since you're probably assuming that the mass is unchanged, [tex] m_\text{new}=m_\text{old} [/tex].

    Then,
    [tex]\begin{align*}
    K_\text{new}&=\frac{1}{2}m_\text{new}v_\text{new}^2\\
    &=\frac{1}{2}(m_\text{old})({\color{red}2}v_\text{old})^2\\
    &=\frac{1}{2}(m_\text{old}){\color{red}2^2}(v_\text{old})^2\\
    &={\color{red}4}\frac{1}{2}(m_\text{old})(v_\text{old})^2\\
    &={\color{red}4} K_\text{old}\\
    \end{align*}
    [/tex]


    You can try this technique to answer your questions.
     
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