The gravitational field in a region is given | vedclass

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(C) Given, gravitational field intensity, $I = (5 \hat{i} + 12 \hat{j}) 	ext{ N kg}^{-1}$.Mass of the object, $m = 3 	ext{ kg}$.The change in gravit

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$48$

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(C) Given, gravitational field intensity, $I = (5 \hat{i} + 12 \hat{j}) 	ext{ N kg}^{-1}$.Mass of the object, $m = 3 	ext{ kg}$.The change in gravitational potential energy $\Delta U$ is given by the negative of the work done by the gravitational field, or $\Delta U = -W = -m \int \vec{I} \cdot d\vec{r}$.However, in many contexts, the magnitude of the change in potential energy is calculated as $\Delta U = m \int_{A}^{B} \vec{I} \cdot d\vec{r}$ considering the work done against the field.Using the formula $\Delta U = m \int_{(0,0)}^{(8,-2)} \vec{I} \cdot d\vec{r}$:$\Delta U = 3 \int_{(0,0)}^{(8,-2)} (5 \hat{i} + 12 \hat{j}) \cdot (dx \hat{i} + dy \hat{j})$$\Delta U = 3 [5x + 12y]_{(0,0)}^{(8,-2)}$$\Delta U = 3 [5(8) + 12(-2)] - 3 [5(0) + 12(0)]$$\Delta U = 3 [40 - 24] = 3 [16] = 48 	ext{ J}$.

The gravitational field in a region is given by $I = (5 \hat{i} + 12 \hat{j}) \text{ N kg}^{-1}$. The change in the gravitational potential energy of an object of mass $3 \text{ kg}$ when it is taken from the origin to a point $(8 \text{ m}, -2 \text{ m})$ is (in $\text{ J}$)

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