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A small block of mass 20 kg rests on a bigger block of mass 30kg , which lies on smooth horizontal plane. Initially the whole system is at rest. The coefficient of friction between the blocks is 0.5 . A horizontal force F=50N is applied on the lower block. Find the work done by frictional force on upper block in t=2sec

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4 years ago

733 Points
```							Here, frictional force = μmg  μ – coeffiecient of friction between the blocks wherm – mass of the small blockg – acceleration due to gravity ∴, F=0.5×20×10=100Now, as the applied force is less than the frictional force there is no relative displacement between the two blocks.So there is no work done by the frictional force.Hence work done by the frictional force in 2 sec = 0. Hope you understand my solution.
```
4 years 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