The displacement-time graph of a particle mov | vedclass

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(D) According to the work-energy theorem, the work done by all forces is equal to the change in kinetic energy: $W = \Delta K = \frac{1}{2} m(v_f^2 -

Vedclass · Accepted Answer

Work done by all forces in region $AB$ is positive.

Vedclass · Answer

(D) According to the work-energy theorem, the work done by all forces is equal to the change in kinetic energy: $W = \Delta K = \frac{1}{2} m(v_f^2 - v_i^2)$.Since the slope of the displacement-time $(s-t)$ graph represents velocity $(v = ds/dt)$, we analyze the slope in each region:$1$. In region $OA$, the slope is constant, so velocity is constant. Thus, $\Delta K = 0$ and work done is zero.$2$. In region $AB$, the slope is increasing, which means the velocity is increasing $(v_f > v_i)$. Therefore, the change in kinetic energy is positive, and the work done by all forces is positive.$3$. In region $BC$, the slope is decreasing, which means the velocity is decreasing $(v_f Thus, the correct statement is that the work done by all forces in region $AB$ is positive.

The displacement-time graph of a particle moving in a straight line is as shown in the figure. Select the correct alternative.

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