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?
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Expert's answer
2012-12-28T07:28:50-0500
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
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