The gravitational field in a region is given by $\vec{E} = (2\hat{i} + 3\hat{j})\ \text{N/kg}$. The amount of work done by the gravitational field when a particle is moved on the line $3y + 2x = 5$ is:

Question

Two identical particles each of mass $ m $ go around a circle of radius $ a $ under the action of their mutual gravitational attraction. The angular speed of each particle will be:

Answer 1

Work by the field is $W = \vec{E}\cdot\vec{d}$, so only the part of the displacement that lies along $\vec{E}$ contributes.

A direction vector for the line $2x + 3y = 5$ can be found from its slope. Rearranged, $y = -\tfrac{2}{3}x + \tfrac{5}{3}$, so a step along the line looks like $\vec{d} = (3,\,-2)$ (moving $+3$ in $x$ drops $-2$ in $y$).

Now dot it with the field:\[\vec{E}\cdot\vec{d} = (2)(3) + (3)(-2) = 6 - 6 = 0.\]The field is at right angles to the line, so no matter how far the particle travels along $3y + 2x = 5$, the field does no work on it.\[\boxed{W = 0}\]

The gravitational field in a region is given by \(\vec{E} = (2\hat{i} + 3\hat{j})\ \text{N/kg}\). The amount of work done by the gravitational field when a particle is moved on the line \(3y + 2x = 5\) is:

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The Correct Option is D

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