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(B) The change in gravitational potential energy $\Delta U$ is given by the work done against the gravitational field, or $\Delta U = -W = -\int \vec{

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$-240$

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(B) The change in gravitational potential energy $\Delta U$ is given by the work done against the gravitational field, or $\Delta U = -W = -\int \vec{F} \cdot d\vec{r} = -m \int \vec{E} \cdot d\vec{r}$.Given $\vec{E} = (5\hat{i} + 12\hat{j}) 	ext{ N/kg}$ and displacement $\vec{r} = (12\hat{i} + 5\hat{j}) 	ext{ m}$.Since the field is uniform, the work done $W = m(\vec{E} \cdot \vec{r})$.$W = 2 	ext{ kg} 	imes [(5\hat{i} + 12\hat{j}) \cdot (12\hat{i} + 5\hat{j})] 	ext{ J}$.$W = 2 	imes (5 	imes 12 + 12 	imes 5) = 2 	imes (60 + 60) = 2 	imes 120 = 240 	ext{ J}$.The change in gravitational potential energy $\Delta U = -W = -240 	ext{ J}$.

The gravitational field in a region is given by the equation $E = (5\hat{i} + 12\hat{j}) \text{ N/kg}$. If a particle of mass $2 \text{ kg}$ is moved from the origin $(0, 0)$ to the point $(12 \text{ m}, 5 \text{ m})$ in this region, the change in gravitational potential energy is: (in $\text{ J}$)

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