A \(1\) kg block subjected to two simultaneous forces \((2\hat{i} + 3\hat{j} + 4\hat{k})\) N and \((3\hat{i} - \hat{j} - 2\hat{k})\) N is moved a distance of \(25\) m along \((3\hat{i} - 4\hat{j})\) direction. The work done in this process is _____ J.

The net force acting on the block is the vector sum of the two forces:

\(\vec{F} = \vec{F}_1 + \vec{F}_2 = (2\hat{i} + 3\hat{j} + 4\hat{k}) + (3\hat{i} - \hat{j} - 2\hat{k}) = 5\hat{i} + 2\hat{j} + 2\hat{k}\) N

The block is moved a distance of \(25\) m along the direction of the vector \(3\hat{i} - 4\hat{j}\). The unit vector along this direction is:

\(\hat{n} = \dfrac{3\hat{i} - 4\hat{j}}{\sqrt{3^2 + (-4)^2}} = \dfrac{3\hat{i} - 4\hat{j}}{5}\)

The displacement vector \(\vec{s}\) is the magnitude multiplied by the unit vector:

\(\vec{s} = 25 \hat{n} = 25 \left( \dfrac{3\hat{i} - 4\hat{j}}{5} \right) = 15\hat{i} - 20\hat{j}\) m

The work done \(W\) is the dot product of the net force and the displacement vector:

\(W = \vec{F} \cdot \vec{s} = (5\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (15\hat{i} - 20\hat{j})\)

\(W = (5 \times 15) + (2 \times -20) + (2 \times 0)\)

\(W = 75 - 40 = 35\) J

Answer: \(35\)