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`        A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The coefficient of friction between the block and the slab is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If g =10 m/s2, the resulting acceleration of the slab will be`
one year ago

Arun
23526 Points
```							Dear student Normal reaction from 40 kg slab on 10 kg block = 10 * 9.81 = 98.1 N Static frictional force = 98.1 * 0.6 N is less than 100 N applied => 10 kg blck will slide on 40 kg slab and net force on it = 100 N - kinetic friction = 100 - 98.1 * 0.4 = 61 N => 10 kg block will slide on 40 kg slab with 61/10 = 6.1 m/s  Frictional force on 40 kg slab by 10 kg block = 98.1 * 0.4 = 39 N => 40 kg slab will move with 39/40 m/s = 0.98 m/s.
```
one year 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