The potential energy of a `1 kg` particle free to move along the x- axis is given by `V(x) =

The potential energy of a `1 kg` particle free to move along the x- axis is given by `V(x) = ((x^(4))/(4) - x^(2)/(2)) J` The total mechainical energy of the particle is `2 J` . Then , the maximum speed (in m//s) is

[@kapilkumar-ms3ns]

Well explained sir

[@zahergaming8280]

Great explanation sir ji👍

[@harshmunjapara3165]

Thank you ❤

[@Legendaxboy]

Thank you sir 🎉🎉😊😊

[@IshuSingh-m7u]

x=0 bhi ayega wo kyon nhi lia h

[@LostUfo669]

Better than peedabloo

[@million-jv8dy]

Not in teaching but in solution