# Answer to Question #21238 in Mechanics | Relativity for swarup

Question #21238

the potential energy of a particle of mass 1kg moving along x axis given by U(x)=[(X)sq/2 -x]joule .if total mechanical energy of particle is 2joule,find its maximum speed?

Expert's answer

E=T+U

T=mx'^2/2

U=x^2/2-x

E=mx'^2/2+x^2/2-x

2=1*x'^2/2

Maximum speed is at point where x''=0

So, we have to solve next differential equation:

x'^2/2+x^2/2-x=2.

x'^2+x^2-2x=4

x'=Sqrt(-x^2+2x-4)

x''=(1-x)/Sqrt(-x^2+2x-4)

So, maximum we have at point x=1

At x=1 x'^2/2+1-2=4

x'^2/2=5

x'^2=10

x'=5=V - maximum speed

T=mx'^2/2

U=x^2/2-x

E=mx'^2/2+x^2/2-x

2=1*x'^2/2

Maximum speed is at point where x''=0

So, we have to solve next differential equation:

x'^2/2+x^2/2-x=2.

x'^2+x^2-2x=4

x'=Sqrt(-x^2+2x-4)

x''=(1-x)/Sqrt(-x^2+2x-4)

So, maximum we have at point x=1

At x=1 x'^2/2+1-2=4

x'^2/2=5

x'^2=10

x'=5=V - maximum speed

## Comments

## Leave a comment