double KE, what happens to velocity

[justinbaker]

if you double the KE, by what factor does the speed increase? and vice versa

i am a little lost

[Pyrrhus]

The speed will increase by \sqrt(2)

Use the equation

K = \frac{1}{2}mv^2

If the speed increases twice

K = \frac{1}{2}m(2v)^2

K = 4 (\frac{1}{2}mv^2)

K = 4K

Kinetic energy will increase four times.

Now if the kinetic energy increases four times

v = \sqrt{\frac{2K}{m}}

v =\sqrt{\frac{(4)2K}{m}}

v = \sqrt{4} \sqrt{\frac{2K}{m}}

v = 2 \sqrt{\frac{2K}{m}}

v = 2v

The speed will increase two times.

[robphy]

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

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

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

Then,
\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{o ld})^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{o ld})^2\\
&={\color{red}4} K_\text{old}\\
\end{align*}



You can try this technique to answer your questions.