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(A) The conservative force $\overrightarrow{F}$ is related to the potential energy $U$ by the relation $\overrightarrow{F} = -
abla U = -\left( \frac

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$-3K$

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(A) The conservative force $\overrightarrow{F}$ is related to the potential energy $U$ by the relation $\overrightarrow{F} = -
abla U = -\left( \frac{\partial U}{\partial x} \hat{i} + \frac{\partial U}{\partial y} \hat{j} + \frac{\partial U}{\partial z} \hat{k} ight)$.Given $U = K(x + y)$,we calculate the partial derivatives:$\frac{\partial U}{\partial x} = K$ and $\frac{\partial U}{\partial y} = K$.Thus,the force is $\overrightarrow{F} = -(K \hat{i} + K \hat{j})$.The displacement vector $\overrightarrow{S}$ from $(1, 1)$ to $(2, 3)$ is $\overrightarrow{S} = (2 - 1) \hat{i} + (3 - 1) \hat{j} = \hat{i} + 2 \hat{j}$.The work done $W$ by the conservative force is given by $W = \overrightarrow{F} \cdot \overrightarrow{S}$.$W = -(K \hat{i} + K \hat{j}) \cdot (\hat{i} + 2 \hat{j}) = -(K 	imes 1 + K 	imes 2) = -(K + 2K) = -3K$.

If the potential energy for a particle is given by $U = K(x + y)$,where $K$ is a constant,what is the work done by the conservative force in moving a particle from $(1, 1)$ to $(2, 3)$?

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