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