In the figure below, the variation of potential energy of a particle of mass m = 2 kg is represented w.r.t its x-coordinate. The particle moves under the effect of the conservative force along the x-axis. Which of the following statements is incorrect about the particle?

 

If the particle is released at the origin, it will try to go in the direction of force. Here dU/dx is positive and hence force is negative; as a result, it will move towards negative x-axis. 

When the particle is released at $x=2+\mathrm{\Delta }\text{,}$ it will reach the point of least possible potential energy (-15 J) where it will have maximum kinetic energy. 

$\begin{array}{l}\frac{1}{2}{\mathrm{mv}}_{max}^2=25\\ {\mathrm{v}}_{max}=5{\mathrm{ms}}^{-1}\end{array}$

The particle will now perform oscillatory motion with x = 5 as mean position. 

$\begin{array}{l}\text{ In (c), }{\mathrm{E}}_{\mathrm{i}}={\mathrm{u}}_{\mathrm{i}}+{\mathrm{k}}_{\mathrm{i}}=15+6=21\mathrm{J}\\ \text{ At }\mathrm{x}=10,{\mathrm{U}}_{\mathrm{f}}=20\\ {\mathrm{K}}_{\mathrm{f}}=1\ne 0\end{array}$

So the particle crosses x = 10.