A 1 kg block subjected to two simultaneous forces (2 i + 3 j + 4 k...

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: $35$