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`        A suitcase is gently dropped on a conveyor belt moving at 3m/s. If the coefficient of friction between the belt and suitcase is 0.5. Find the displacement of suitcase relative to conveyor belt before the slipping b/w the two is stopped `
3 years ago

Vikas TU
10402 Points
```							Friction Force = umg = mg/2accln. = > g/2 No vertical accln. as converyor belt is moving horizontally.From Newton’s second law of eqn.=>s = 0.5*a*t^2 and0 = u – att = u/a => 3/0.5g => 6/gand distance traveled => 0.5 * g/2 * (6/g)^2 => 0.9 meter approx.
```
3 years ago
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• 141 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions