A body of mass 1kg is moving on the x axis under the action of a force such that its variation of PE as a function of time is given by U=x-x^2.Find its (1)equilibrium position and tell its nature (2)velocity at equilibrium position if its velocity at origin is 2m/s.

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Arun · Answer

Dear student At equilibrium position dU/dx should be zero 1 - 2x = 0 Hence x = 1/2 metres Regards Arun (askIITians forum expert)

prasad · Answer

plz also find the velocity at equlibrium position if its velocity at origin is 2m/s.

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A body of mass 1kg is moving on the x axis under the action of a force such that its variation of PE as a function of time is given by U=x-x^2.Find its (1)equilibrium position and tell its nature (2)velocity at equilibrium position if its velocity at origin is 2m/s.

A body of mass 1kg is moving on the x axis under the action of a force such that its variation of PE as a function of time is given by U=x-x^2.Find its (1)equilibrium position and tell its nature
(2)velocity at equilibrium position if its velocity at origin is 2m/s.

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A body of mass 1kg is moving on the x axis under the action of a force such that its variation of PE as a function of time is given by U=x-x^2.Find its (1)equilibrium position and tell its nature (2)velocity at equilibrium position if its velocity at origin is 2m/s.

A body of mass 1kg is moving on the x axis under the action of a force such that its variation of PE as a function of time is given by U=x-x^2.Find its (1)equilibrium position and tell its nature(2)velocity at equilibrium position if its velocity at origin is 2m/s.

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A body of mass 1kg is moving on the x axis under the action of a force such that its variation of PE as a function of time is given by U=x-x^2.Find its (1)equilibrium position and tell its nature
(2)velocity at equilibrium position if its velocity at origin is 2m/s.