A point particle of mass 0.5 kg is moving al | vedclass

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(B) From the graph, the potential energy $V$ at the origin $(x=0)$ is $0 \ J$. To escape to infinity, the particle must cross the potential energy bar

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$2$

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(B) From the graph, the potential energy $V$ at the origin $(x=0)$ is $0 \ J$. To escape to infinity, the particle must cross the potential energy barrier on either side. The peak of the potential energy barrier on the left side is $V_{left} = 1 \ J$ at $x = -1 \ m$. The peak of the potential energy barrier on the right side is $V_{right} = 4 \ J$ at $x = 3 \ m$. To escape, the particle must have enough kinetic energy to overcome the lower barrier. Therefore, the minimum kinetic energy required is $K_{min} = V_{left} = 1 \ J$. Using the kinetic energy formula $K = \frac{1}{2}mv^2$, we have: $\frac{1}{2} 	imes 0.5 	imes v^2 = 1$ $0.25 	imes v^2 = 1$ $v^2 = 4$ $v = 2 \ ms^{-1}$.

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