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 two simultaneous forces on the block are
$$\mathbf{F}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k} \; \text{N}$$
$$\mathbf{F}_2 = 3\hat{i} - \hat{j} - 2\hat{k} \; \text{N}$$

Step 1 - Find the resultant force.
Add the vectors component-wise:
$$\mathbf{F} = \mathbf{F}_1 + \mathbf{F}_2$$ $$= (2+3)\hat{i} + (3-1)\hat{j} + (4-2)\hat{k}$$ $$= 5\hat{i} + 2\hat{j} + 2\hat{k}\; \text{N}$$

Step 2 - Write the displacement vector.
The block moves 25 m along the direction $$3\hat{i} - 4\hat{j}$$.

First find the unit vector in that direction.
Magnitude of $$3\hat{i} - 4\hat{j}$$ is
$$|\;3\hat{i} - 4\hat{j}\;| = \sqrt{3^{2} + (-4)^{2}} = \sqrt{9 + 16} = 5$$

Hence the unit vector is
$$\hat{u} = \frac{3}{5}\hat{i} - \frac{4}{5}\hat{j}$$

Multiply the unit vector by the distance 25 m to get the displacement vector $$\mathbf{s}$$:
$$\mathbf{s} = 25\hat{u} = 25\left(\frac{3}{5}\hat{i} - \frac{4}{5}\hat{j}\right)$$ $$= 15\hat{i} - 20\hat{j} + 0\hat{k}\; \text{m}$$

Step 3 - Calculate the work done.
Work $$W$$ is the dot product of the resultant force and displacement:
$$W = \mathbf{F} \cdot \mathbf{s}$$ $$= (5\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (15\hat{i} - 20\hat{j} + 0\hat{k})$$ $$= 5 \times 15 + 2 \times (-20) + 2 \times 0$$ $$= 75 - 40 + 0$$ $$= 35 \; \text{J}$$

Therefore, the work done in moving the block is 35 J.