A $$40$$ kg slab rests on a frictionless floor. A $$l0$$ 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

Question

Infinity Learn · Accepted Answer

The maximum force for which there will be no relative motion between the blocks is given by   Fmax $$=$$ μmg[$$1+mM$$] $$=$$ $$0.6$$×$$10$$×$$10$$ [$$1+1040$$]N $$=75$$ N Since the applied force (100N) is greater than Fmax, the $$10$$ kg block will slip over $$40$$ kg block. Therfore acceleration of $$10$$ kg block $$=$$ $$100-0.4$$×$$10$$×$$1010=6$$ $$m/s2$$

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