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

Question

Infinity Learn · Accepted Answer

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

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