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A particle, which is constrained to move along the x-axis, is subjected to a force from the origin as `F(x) = -kx + ax^(3)`.Here `k` and a are positive constants. For `x=0`, the functional form of the potential energy `U(x)` of particle is.A. (a) B. (b) C. (c) D. (d) |
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Answer» Correct Answer - D `dU_((x))=-Fdx` `:. U_x=underset0oversetx intFdx=(kx^2)/(2)-(ax^4)/(4)` `U=0` at `x=0` and at `x=sqrt((2k)/(a))` `U(x)` is negative for `xgtsqrt((2k)/(a))` `F=0` at `x=0`. It means slope of `U-x` graph is zero at `x=0`. |
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