The potential energy of a 1 kg particle free to move along the x-axis is given by V(x) ((x^4)/4 - (x^2)/2)J The total mechanical energy of the particle is 2 J. Then, the maximum speed (in m/s) is

The potential energy of a 1 kg particle free to move along the x-axis

1.	The potential energy of a 1 kg particle free to move along the x-axis is given by V(x) ((x^4)/4 - (x^2)/2)J The total mechanical energy of the particle is 2 J. Then, the maximum speed (in m/s) is
Answer» 2 `3/(sqrt2)` `sqrt(2)` `1/(sqrt2)` Solution :Total energy `E_t = 2J` . It is fixed . For maximum SPEED. KINETIC energy is maximum. The potential energy should therefore be minimum. As `V (x) = (x^4)/(4) - (x^2)/(2)` `:. (DV)/(dx) = (4x^3)/(4) - (2x)/(2) = x^3 - x = x(x^2 -1)` For V to be minimum, `(dV)/(dx) = 0` `:. x(x^2 - 1) = 0 , "or " x = 0, pm 1` At `x = 0, V(x)= 0` and at `x = pm 1, V(x) = -1/4 J` `:. ("Kinetic energy")_("max") = E_T - V_("min")` or `("Kinetic energy")_("max")= 2 - (-1/4) = 9/4 J`. or `1/2 mv_("max")^(2) = 9/4"or " v_("max")^(2) = (9 xx 2)/(m xx 4) = (9 xx 2)/(1 xx 4) = 9/2` `:. v_("max") = 3/(sqrt2) m//s` .