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Q.

A 40-kg slab rests on frictionless floor as shown in the figure. 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. A horizontal force of 100 N acts upon the 10 kg block. The resulting acceleration of the slab will be

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a

1 ms−2

b

1.5 ms−2

c

2 ms−2

d

2.5 ms−2

answer is A.

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Detailed Solution

fL=μsmg=0.6×10×10=60NF>fLDue to the applied force the block moves kinetic friction moves the slabfor the slab F=fk=μkmg ⇒Ma=0.4×10×10 40×a=0.4×100 ⇒a=1ms−2
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A 40-kg slab rests on frictionless floor as shown in the figure. 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. A horizontal force of 100 N acts upon the 10 kg block. The resulting acceleration of the slab will be