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 \[=\,\,\,{{\mu }_{s}}{{m}_{A}}g=\,\,0.6\times 10\times 9.8=58.8\,\,N\] 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 \[{{\operatorname{F}}_{k}}=\,\,{{\mu }_{k}}{{m}_{A}}g\,=\,\,0.4\times 10\times 9.8\,\,=\,\,39.2\,\,N\] This kinetic friction helps to move the slab \[\therefore Accleration of slab = \frac{39.2}{{{m}_{B}}}=\frac{39.2}{40}=0.98\,\,m/{{s}^{2}}\]You need to login to perform this action.
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