Questions
The potential energy of a 1 kg particle free to move along the x-axis is given by Joules. The total Mechanical Energy of the particle 2J. Then the maximum speed is,
detailed solution
Correct option is B
total energy TE=2 Jfor maximum speed⇒KE is maximum, ⇒PE is minimumPE=V(x)=x44-x22 differentiate with respect to x , to find minimum PEdVdx=4x34-2x2dVdx=x3-xfor maximum or miinimum dVdx=0x3-x=0 ⇒x2=1 ⇒x=±1for minimum value of PE consider x=-1V(x)=x44-x22substitute x=-1V(x)=-14=PE minimumTE=KE+PEKEmax=TE-PEminKEmax=2--14KEmax=9412mv2=94v=32m/sTalk to our academic expert!
Similar Questions
A particle free to move along the x-axis has potential energy given by for , where k is a positive constant of appropriate dimensions. Then
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