A particle of mass m is constrained to move o | vedclass

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(A) When the particle is to the left of $E$,the force $F$ is in the right direction,which is positive. Since the magnitude of $F$ is a constant $F_0$,

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Vedclass · Answer

(A) When the particle is to the left of $E$,the force $F$ is in the right direction,which is positive. Since the magnitude of $F$ is a constant $F_0$,the acceleration $a = F_0 / m$ is a positive constant. Since $a = dv/dt$,the slope of the $v-t$ graph is a positive constant. Thus,the velocity increases linearly with time until the particle reaches $E$. After the particle passes $E$,the force $F$ acts in the left direction,which is negative. Thus,the acceleration becomes a negative constant,and the slope of the $v-t$ graph is a negative constant. The velocity decreases linearly,becomes zero,and then becomes negative as the particle moves to the right of $E$. This motion repeats,resulting in a triangular wave pattern for the velocity-time graph. Therefore,the correct graph is the one showing a linear increase and decrease in velocity,which corresponds to option $A$.

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