A) \[0.98\,m/{{s}^{2}}\]
B) \[1.47\,m/{{s}^{2}}\]
C) \[1.52\,m/{{s}^{2}}\]
D) \[6.1\,m/{{s}^{2}}\]
Correct Answer: A
Solution :
Limiting friction between block and slab\[={{\mu }_{s}}{{m}_{A}}g\] \[=0.6\times 10\times 9.8=58.8N\] But applied force on block A is 100 N. So the block will slip over a slab. Now kinetic friction works between block and slab \[{{F}_{k}}={{\mu }_{k}}{{m}_{A}}g\] \[=0.4\times 10\times 9.8=39.2\ N\] This kinetic friction helps to move the slab \[\therefore \] Acceleration of slab\[=\frac{39.2}{{{m}_{B}}}=\frac{39.2}{40}=0.98\ m/{{s}^{2}}\]
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