- #1
autodidude
- 333
- 0
In K&K's text on mechanics, after they present the derivation of the work energy theorem:
[tex]\frac{1}{2}mv^2-\frac{1}{2}mv_0^2=\int_x_0^x F(x) dx[/tex]
It is mentioned that since [tex]v=\frac{dx}{dt}[/tex], we could solve for [tex]\frac{dx}{dt}[/tex] and integrate again to find x(t)
I tried that with [tex]v_0=0[/tex] just to make things a little easier and ended up with something like:
[tex]x=\int \sqrt{2m(F(x)-F(x_0))}dt[/tex]
which looks horrible and wrong
Is this heading down the wrong path? How would you do it when [tex]v_0[/tex] is not zero?
[tex]\frac{1}{2}mv^2-\frac{1}{2}mv_0^2=\int_x_0^x F(x) dx[/tex]
It is mentioned that since [tex]v=\frac{dx}{dt}[/tex], we could solve for [tex]\frac{dx}{dt}[/tex] and integrate again to find x(t)
I tried that with [tex]v_0=0[/tex] just to make things a little easier and ended up with something like:
[tex]x=\int \sqrt{2m(F(x)-F(x_0))}dt[/tex]
which looks horrible and wrong
Is this heading down the wrong path? How would you do it when [tex]v_0[/tex] is not zero?