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`        A BLOCK OF MASS 2 KG IS PLACED ON THE FLOOR. THE COEFFICIENT OF STATIC FRICTION IS 0.4. IF  A FORCE OF 1.25 ACTS FOR 8 SEC. FINAL VELOCITY OF THE BODY IS`
2 years ago

Arun
23381 Points
```							We have, coefficient of static friction, µs = 0.4Mass, m = 2 kgForce applied, F = 2.5 NMaximum static frictional force, fmax= µs × m × g = 0.4 × 2 × 9.8=> fmax = 7.84 NSince, the applied force is less than the maximum static frictional force, the frictional force on the block is equal to the applied force = 2.5 N. This is according to the fact that static friction is a self adjusting force.
```
2 years ago
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• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions