JEE_MAINS_2026_Physics-Apr4-Shit1_Q46_Work_Done_By_Forces #jeemain2026

JEE MAINS 2026 Physics Question 46 - Complete Visual Solution!
In this video, we solve JEE MAINS 2026 Physics Q46 on Work Done by Two Forces on a block with detailed visual explanations using animations.
Topics Covered:
- Work done by a force: W = F . d
- Vector dot product calculation
- Net force from multiple forces
- Displacement along a given direction
Given: F1 = (2i + 3j + 4k) N, F2 = (3i - j - 2k) N, mass = 1 kg. Displacement: 25 m along (3i - 4j) direction.
Unit vector along displacement: (3i - 4j)/5.
Displacement vector d = 25 * (3i - 4j)/5 = (15i - 20j) m.
Net force F = F1 + F2 = (5i + 2j + 2k) N.
Work W = F . d = 5(15) + 2(-20) + 2(0) = 75 - 40 = 35 J.
Answer: 35 J
JEE MAINS 2026 | PHYSICS | Work Energy Power | Dot Product
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Problem Walkthrough:
Step 1: Read the question - Two forces F1=(2i+3j+4k)N and F2=(3i-j-2k)N act on a 1 kg block. It moves 25 m along (3i-4j). Find work done.
Step 2: Find net force: F = F1 + F2 = (2+3)i + (3-1)j + (4-2)k = (5i+2j+2k) N.
Step 3: Find displacement vector. Direction = (3i-4j), magnitude = sqrt(9+16) = 5.
Step 4: Unit vector = (3i-4j)/5. Displacement d = 25*(3i-4j)/5 = (15i-20j) m.
Step 5: Calculate work: W = F . d = 5*15 + 2*(-20) + 2*0 = 75 - 40 + 0 = 35 J.
Step 6: Answer is 35 J. Key: Add forces vectorially first, then compute dot product with displacement vector.
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