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(C) According to the Work-Energy Theorem,the change in kinetic energy $(\Delta KE)$ is equal to the work done $(W)$ by the force on the particle.The w

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$25$

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(C) According to the Work-Energy Theorem,the change in kinetic energy $(\Delta KE)$ is equal to the work done $(W)$ by the force on the particle.The work done is given by the line integral: $W = \int \overrightarrow{F} \cdot d\overrightarrow{r} = \int (4x \hat{i} + 3y^2 \hat{j}) \cdot (dx \hat{i} + dy \hat{j})$.This simplifies to: $W = \int_{1}^{2} 4x \, dx + \int_{2}^{3} 3y^2 \, dy$.Evaluating the integrals:$W = [2x^2]_{1}^{2} + [y^3]_{2}^{3}$.$W = (2(2)^2 - 2(1)^2) + (3^3 - 2^3)$.$W = (8 - 2) + (27 - 8)$.$W = 6 + 19 = 25 \, J$.Therefore,the kinetic energy changes by $25 \, J$.

$A$ particle experiences a variable force $\overrightarrow{F} = (4x \hat{i} + 3y^2 \hat{j})$ in a horizontal $x-y$ plane. Assume distance is in meters and force is in newtons. If the particle moves from point $(1, 2)$ to point $(2, 3)$ in the $x-y$ plane,the kinetic energy changes by............$J$.

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