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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

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a

1.5 ms-2

b

6.0 ms-2

c

10 ms-2

d

1.0 ms-2

answer is B.

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Detailed Solution

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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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