Q.

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

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a

0.98 m/s2

b

1.47 m/s2

c

1.52 m/s2

d

6.1 m/s2

answer is A.

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Detailed Solution

The frictional force between 10 kg block and slab of 40 kg is given by f=μmg=0⋅4×10×9⋅8N.The acceleration of the slab of 40 kg is a=0⋅4×10×9⋅840=0⋅98m/s2.
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