Question

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:

• A. \(4\)
• B. \(30\)
• C. \(25\)
• D. \(0\)

Hint:

Compare the field vector \((2,3)\) with the line's direction. If they are perpendicular, \(\vec{E}\cdot\vec{d}=0\).

Answer 1

The Correct Option is D

Step 1: The work done per unit mass by the field over a displacement \(\vec{d}\) is \(W = \vec{E}\cdot\vec{d}\). Here \(\vec{E} = 2\hat{i} + 3\hat{j}\).

Step 2: The particle moves along the line \(2x + 3y = 5\). Writing this as \(2x + 3y = 5\), the coefficients \((2,3)\) form the normal vector to the line.

Step 3: Notice that the field \(\vec{E} = (2,3)\) is exactly the same as this normal direction. So the field points perpendicular to the line, meaning it is perpendicular to every displacement made along the line.

Step 4: Since the displacement along the line is perpendicular to \(\vec{E}\), the dot product is zero:\[W = \vec{E}\cdot\vec{d} = |\vec{E}||\vec{d}|\cos 90^\circ = 0.\]This is option (D).\[\boxed{W = 0}\]