Work Done by Simultaneous Forces on Moving Block

To find the total work done on the block, we first determine the net force acting on it.

The two given forces are: $\vec{F}_1 = (2\hat{i} + 3\hat{j} + 4\hat{k})\text{ N}$ $\vec{F}_2 = (3\hat{i} - \hat{j} - 2\hat{k})\text{ N}$

The net force $\vec{F}_{\text{net}}$ acting on the block is the vector sum of these two forces: $\vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2 = (2 + 3)\hat{i} + (3 - 1)\hat{j} + (4 - 2)\hat{k}$ $\vec{F}_{\text{net}} = (5\hat{i} + 2\hat{j} + 2\hat{k})\text{ N}$

Next, we find the displacement vector $\vec{s}$. The direction of motion is given by the vector $\vec{d} = 3\hat{i} - 4\hat{j}$. The unit vector $\hat{d}$ in this direction is: $\hat{d} = \frac{3\hat{i} - 4\hat{j}}{\sqrt{3^2 + (-4)^2}} = \frac{3\hat{i} - 4\hat{j}}{\sqrt{9 + 16}} = \frac{3\hat{i} - 4\hat{j}}{5}$

Given that the block moves a distance of $25\text{ m}$ in this direction, the displacement vector $\vec{s}$ is: $\vec{s} = 25 \cdot \hat{d} = 25 \left( \frac{3\hat{i} - 4\hat{j}}{5} \right) = 5(3\hat{i} - 4\hat{j}) = (15\hat{i} - 20\hat{j})\text{ m}$

The work done $W$ by the simultaneous forces is given by the scalar product of the net force and the displacement vector: $W = \vec{F}_{\text{net}} \cdot \vec{s}$ $W = (5\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (15\hat{i} - 20\hat{j} + 0\hat{k})$ $W = (5)(15) + (2)(-20) + (2)(0)$ $W = 75 - 40 + 0 = 35\text{ J}$

Thus, the work done in this process is $35\text{ J}$.