The adjoining diagram shows the velocity vers | vedclass

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(A) According to the work-energy theorem, the work done by the net force on a particle is equal to the change in its kinetic energy, $W = \Delta K = \

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$A$ to $B$

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(A) According to the work-energy theorem, the work done by the net force on a particle is equal to the change in its kinetic energy, $W = \Delta K = \frac{1}{2}m(v_f^2 - v_i^2)$.For the work done to be positive, the final kinetic energy must be greater than the initial kinetic energy, which implies that the final velocity $v_f$ must be greater than the initial velocity $v_i$.Looking at the velocity-time graph:$1$. In the region $AB$, the velocity increases with time, so $v_f > v_i$. Thus, the work done is positive.$2$. In the region $BC$, the velocity is constant, so $v_f = v_i$. Thus, the work done is zero.$3$. In the region $CD$, the velocity decreases, so $v_f $4$. In the region $DE$, the velocity is constant, so $v_f = v_i$. Thus, the work done is zero.Therefore, the work done by the force on the particle is positive from $A$ to $B$.

The adjoining diagram shows the velocity versus time plot for a particle. The work done by the force on the particle is positive from

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