A $$40kg$$ slab rests on a frictionless floor as shown in the figure. A $$10kg$$ block rests on the top of the slab. The static coefficient of friction between the block and slab is $$0.60$$ , while the coefficient of kinetic friction is $$0.40$$. The $$10kg$$ block is acted upon by a horizontal force $$100N$$. If $$g=9.8ms$$−$$2$$, the resulting acceleration of the slab will be

Question

A $$40kg$$ slab rests on a frictionless floor as shown in the figure. A $$10kg$$ block rests on the top of the slab. The static coefficient of friction between the block and slab is $$0.60$$ , while the coefficient of kinetic friction is $$0.40$$. The $$10kg$$ block is acted upon by a horizontal force $$100N$$. If $$g=9.8ms$$−$$2$$, the resulting acceleration of the slab will be

Infinity Learn · Accepted Answer

Limiting friction between block and slab $$=$$μ$$smAg=0.6$$×$$10$$×$$9.8=58.8$$ NBut applied force on block A is $$100$$ N.So the block will slip over the slab.Now kinetic friction works between block and $$slabFk=$$μ$$kmAg=0.4$$×$$10$$×$$9.8=39.2$$ NThis kinetic friction helps to move the slab∴Accleration of $$slab=39.2$$ $$mB=39.240=0.98$$ $$m/s2$$

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