• question_answer A particle of mass 1 kg is released from rest at origin to move ina conservative force field. The variation of potential energy corresponding to this conservative force with X-coordinate is as shown in the figure. Mark out the correct statement(s) for this situation. A)  Particle is moving along +ve X-axis                                     B)  Particle may move along+ve X-axis                                  C)  Particle may move along -ve X-axis D)  When particle crosses x =-2 m, the speed of particle is $\sqrt{60}\,m/s$

Here the information about variation of field is given by along X-axis but about $Y$ and $Z$-axis no information is given. On $Y$ and $Z$-coordinate, U may be depending or not. If $\frac{\partial U}{\partial y}=\frac{\partial U}{az}=0,$, then motion of particle is determined only by From graph, ${{F}_{x}}=-\frac{\partial U}{\partial x}=-15\,N,\,-ve$ sign shows that ${{F}_{x}}$ is along $-ve$ X-axis. Consider two cases ; Case I $\frac{\partial U}{\partial y}=\frac{\partial U}{\partial z}=0,$ then particle is going to move along -ve X-axis with an acceleration of $15\,\,m/{{s}^{2}}$ (towards -ve X). Case II $\frac{\partial U}{\partial y}$ and/or $\frac{\partial U}{\partial z}\ne 0,$. then particle's motion cannot be exactly determined from given information but it is for sure that in any circumstance, the particle won't move along +ve X-axis, So, particle may move in -ve X-axis. From work energy theorem $dK=-dU$ $\Rightarrow$  $\frac{m{{v}^{2}}}{2}-0=-\{[-30+{{U}_{y}}+{{U}_{z}}]\,-{{[u]}_{0,\,0,\,0}}\}$ $v=\sqrt{60}\,m/s$ if U is not depending on y and z.
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