JEE Main 2026 — Work Power Energy Question with Solution

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:

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

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

$\vec{s}=25\hat{n}=25\left(\frac{3\hat{i}-4\hat{j}}{5}\right)=15\hat{i}-20\hat{j}$ 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: $35$