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Grade: 12th pass
        
A 40kg slab (B) rests on a smooth floor. A 10kg block (A) rests on the top of the slab. The static coefficient of friction between slab and block is 0.6 while the kinetic friction coefficient is 0.4.  The block A is acted upon by a horizontal force 100N. If g=9.8m/s^2, the resulting acceleration of slab B will be
6 months ago

Answers : (1)

Arun
21041 Points
							
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.
 
6 months ago
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