If there was a graphical representation, I would be happy to assist you.
Answer: ![v=\sqrt[]{\frac{2K}{m} }](https://tex.z-dn.net/?f=v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D)
Step-by-step explanation:

First, multiply by 2 to get rid of the 2 in the denominator. Remember that if you make any changes you have to make sure the equation keeps balanced, so do it on both sides as following;


Divide by m to isolate
.


To eliminate the square and isolate v, extract the square root.
![\sqrt[]{\frac{2K}{m} }=\sqrt[]{v^2}](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D%3D%5Csqrt%5B%5D%7Bv%5E2%7D)
![\sqrt[]{\frac{2K}{m} }=v](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D%3Dv)
let's rewrite it in a way that v is in the left side.
![v=\sqrt[]{\frac{2K}{m} }](https://tex.z-dn.net/?f=v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D)
Given
P(1,-3); P'(-3,1)
Q(3,-2);Q'(-2,3)
R(3,-3);R'(-3,3)
S(2,-4);S'(-4,2)
By observing the relationship between P and P', Q and Q',.... we note that
(x,y)->(y,x) which corresponds to a single reflection about the line y=x.
Alternatively, the same result may be obtained by first reflecting about the x-axis, then a positive (clockwise) rotation of 90 degrees, as follows:
Sx(x,y)->(x,-y) [ reflection about x-axis ]
R90(x,y)->(-y,x) [ positive rotation of 90 degrees ]
combined or composite transformation
R90. Sx (x,y)-> R90(x,-y) -> (y,x)
Similarly similar composite transformation may be obtained by a reflection about the y-axis, followed by a rotation of -90 (or 270) degrees, as follows:
Sy(x,y)->(-x,y)
R270(x,y)->(y,-x)
=>
R270.Sy(x,y)->R270(-x,y)->(y,x)
So in summary, three ways have been presented to make the required transformation, two of which are composite transformations (sequence).