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Question

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on the top of the slab as shown in fig. The static coefficient of friction between the block and the slab is 0.60 while the kinetic coefficient is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If g = 9.8 m s2, the resulting acceleration of the slab will be
1032456_d9c386a03e8a4c91b6238231d1e7a0c7.jpg

A
0.784ms2
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B
1.52ms2
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C
1.47ms2
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D
0.98ms2
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Solution

The correct option is A 0.784ms2
μs=0.6(fs)max=0.6×10×18=58.8Nμk=0.4fk=0.4×10×9.8=39.2Nso,100N>58.8N
Hence 10kg will move and surface will create fk force on 10kg in right direction and hence fk force will be introduced on 40kg slab in left direction.
Net force on slab =fk=39.2N
aslab=39.210+40=0.784ms2


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