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

k E m a x = E T − U m i n U m i n ( ± 1 ) = − 1/4 J KE m a x = 9/4 J ⇒ U = 2 3 J

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Physics - W.E.P. Question with Solution | TestHub

PhysicsW.E.P.Conservative force & potential energyHard2 minQB

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

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Answer:

Solution:

$k E_{m a x} = E_{T} - U_{m i n}$ $U_{m i n} (\pm 1) = - 1/4 J$ $KE_{m a x} = 9/4 J \Rightarrow U = \frac{3}{2} J$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Conservative force & potential energy

⏱ 2mℹ️ Source: QB

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