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`        Three blocks A,B,C of masses 4,2,1kg respectively are in contact on a frictionless surface. If a force of 14N is applied first block A, then the contact force between A and B is what? `
10 months ago

Arun
22572 Points
```							 the acceleration due to force 14N= F=maF/m =a14/(4+2+1)14/7=2m/s2 the contact force between A and B is used to move the bodies B and C only.F=maF=(2+1)2F=3×2=6N
```
10 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions