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The potential energt of a particle of mass 0.1 kg, moving along the x-axis, is given by `U=5x(x-4)J`, where x is in meter. It can be concluded thatA. The particle is acted upon by a variable force.B. The minimum potential energy during motion is `-20 J`.C. The speed of the particle is maximum at `x = 2m`.D. `x = 2` is position of ustable equilibrium |
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Answer» Correct Answer - A::B::C `U = 5x^(2) - 20x` `F = -(dU)/(dx) -[10x - 20]` `:. X = 2` is position of equilibrium Now `(d^(2)U)/(dx^(2)) = 10 gt 0` `:.` Only one minima `rArr x = 2` is minima `rArr KE` is maximum at `x = 2` `U(x = 2) = -20 J` |
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