Question

# A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The static coefficient of friction between the block and the slab is 0.60 while the kinetic coefficient of friction is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N . The resulting acceleration of the slab will be

A
1.5 ms2
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B
2.0 ms2
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C
10 ms2
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D
1.0 ms2
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Solution

## The correct option is D 1.0 ms−2Considering block of mass 10kg N=mg=10×10=100 N So to get static friction fs=μsN=0.6×100=60 N and friction on this block will be in opposite direction of the force applied. So net force on 10 kg block apart from friction is Fnet=100 N So if net force is greater than static friction, then it is a case of kinetic friction. So, fk=μkN⇒fk=0.4×100=40 N Now taking 40 kg block into consideration As fk was acting in opposite direction of force for 10 kg block, it will act in reverse direction for 40 kg block. So only force acting on 40 kg block will be fk ⇒ma=fk⇒40a=40⇒a=1 ms−2

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