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

The correct answer is Option 35. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q46 – JEE-MAIN 2026

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Home Physics Work, Energy & PowerQuestion #646

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.

Enter Calculated Numerical Value:

Clue 1 of 3: Net Force and Work

The work done by multiple forces is the work done by the net force. Work is a scalar product of force and displacement.

🧠 Full Solution Path

Step 1: Calculate the Net Force✦ Active

The net force ${\vec{F}}_{n e t}$ is the vector sum of the two simultaneous forces ${\vec{F}}_{1}$ and ${\vec{F}}_{2}$ .

{\vec{F}}_{n e t} = {\vec{F}}_{1} + {\vec{F}}_{2} = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) + (3 \hat{i} - \hat{j} - 2 \hat{k})

{\vec{F}}_{n e t} = (2 + 3) \hat{i} + (3 - 1) \hat{j} + (4 - 2) \hat{k}

{\vec{F}}_{n e t} = (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) N

Step 2: Determine the Displacement Vector○ Expand

The block is moved a distance of $25 m$ along the direction $(3 \hat{i} - 4 \hat{j})$ . First, find the unit vector in this direction, then multiply by the magnitude of displacement.

Direction vector {\vec{r}}_{d i r} = 3 \hat{i} - 4 \hat{j}

Magnitude of direction vector | {\vec{r}}_{d i r} | = \sqrt{3^{2} + (- 4)^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5

Unit vector \hat{d} = \frac{3 \hat{i} - 4 \hat{j}}{5}

Displacement vector \vec{d} = (25 m) \times \hat{d} = 25 (\frac{3 \hat{i} - 4 \hat{j}}{5})

\vec{d} = 5 (3 \hat{i} - 4 \hat{j}) = (15 \hat{i} - 20 \hat{j}) m

💡 Teacher's Secret Hint

Ensure the displacement vector is correctly formed using the given magnitude and direction.

Step 3: Calculate Work Done○ Expand

Work done $W$ is the dot product of the net force ${\vec{F}}_{n e t}$ and the displacement vector $\vec{d}$ .

W = {\vec{F}}_{n e t} \cdot \vec{d}

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

W = 35 J

💡 Teacher's Secret Hint

Remember that the dot product of orthogonal unit vectors is zero (e.g., $\hat{i} \cdot \hat{j} = 0$ ), and for parallel unit vectors it is one (e.g., $\hat{i} \cdot \hat{i} = 1$ ).

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Physics Question 46 – JEE-MAIN 2026

Clue 1 of 3: Net Force and Work

🧠 Full Solution Path