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The potential energy of a 1kg particle free to move along the x-axis is given by `V(x)=(x^(4)/4-x^(2)/2)J` The total mechanical energy of the particle is 2J then the maximum speed `(in m//s)` isA. `3/sqrt(2)`B. `3sqrt(2)`C. `9/2`D. `2` |
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Answer» Correct Answer - A `V(x) = (x^(4)/4-x^(2)/2)` So when pe is minimum then ke is maximum ` or d/dx(v(x))=0Rightarrow x^(3)-x=0` `Rightarrow x(x^(2)-1)=0` ` Rightarrowx=0,pm1` `V(0)=0(maximum)` ` V(pm1)=-1/4(minimum)` `RightarrowKE_(max)=TE-PE_(min)` `1/2 mV_(max)^(2)=2-(-1/4)=9/4` `V_(max)=sqrt(2(9//4))/(1kg)=3/sqrt(2)m//s` Ans |
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