A particle free to move along the `X-`axis has potential energy given by `U(X)=k[1-exp(-X^(2))]` of appropriate dimensions. Then-

A particle free to move along the (x - axis) has potential energy given by U(x)= k[1 - exp(-x^2)] for -o o le x le + o o , where (k) is a positive constant of appropriate dimensions. Then.

A particle free to move along the x- axis has potential energy given by U(x) = k[1 - e^(-x^(2))] for -oo le x le + oo , where k is a positive constant of appropriate dimensions. Then select the incorrect option

A particle free to move along the x- axis has potential energy given by U(x) = k[1 - e^(-x^(2))] for -oo le x le + oo , where k is a positive constant of appropriate dimensions. Then select the incorrect option

A particle free to move along the (x - axis) hsd potential energy given by U(x)= k[1 - exp(-x^2)] for -o o le x le + o o , where (k) is a positive constant of appropriate dimensions. Then.

A particle free to move along the (x - axis) hsd potential energy given by U(x)= k[1 - exp(-x^2)] for -o o le x le + o o , where (k) is a positive constant of appropriate dimensions. Then.

A particle free to move along the (x - axis) hsd potential energy given by U(x)= k[1 - exp(-x^2)] for -o o le x le + o o , where (k) is a positive constant of appropriate dimensions. Then.

A single conservative fore F_(x) acts on a 2kg particle that moves along the x-axis. The potential energy is given by. U =(x-4)^(2)-16 Here, x is in metre and U in joule. At x =6.0 m kinetic energy of particle is 8 J find . (a) total mechanical energy (b) maximum kinetic energy (c) values of x between which particle moves (d) the equation of F_(x) as a funcation is zero (e) the value of x at which F_(x) is zero .