The potential energy of a 2kg particle is given by U=(ax+by)J, Where a and b constants, x and y are coordinates of particle in metre. Initially the particle is at rest and at (4a,4b).The coordinates after 2 seconds is (xa,xb), where x is______

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answer is 3.

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Detailed Solution

F→=−∂U∂xi^−∂U∂yj^F→=−ai^−bj^ax=Fxm=−−a2;ay=Fym=−b2Sx=12axt2=−12×a2×4=−aSy=12ayt2=12×−b24=−b∴ Find coordinates =[(4a−a),(4b−b)]=(3a,3b)

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