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Grade: 11
        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
3 years ago

Answers : (1)

Adarsh
733 Points
							
Here, frictional force = μmg 
 μ – coeffiecient of friction between the blocks wher
m – mass of the small block
g – acceleration due to gravity
 ∴, F=0.5×20×10=100
Now, 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.
3 years ago
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