A) \[1\,m/{{s}^{2}}\]
B) \[1.5\,m/{{s}^{2}}\]
C) \[2\,m/{{s}^{2}}\]
D) \[6\,m/{{s}^{2}}\]
Correct Answer: A
Solution :
Limiting friction between block and slab \[={{m}_{S}}{{m}_{A}}g=0.6\times 10\times 1=60N\] But applied force on block A is 100 N. So the block will slip over the slab. Now, kinetic friction works between block and slab, \[{{F}_{k}}={{m}_{k}}{{m}_{A}}g=0.4\times 10\times 10=40N\] This kinetic friction helps to move the slab. \[\therefore \]Acceleration of slab \[=\frac{40}{m{{ & }_{B}}}=\frac{40}{40}=1\,m/{{s}^{2}}\]You need to login to perform this action.
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