A) 1150 N
B) 1250 N
C) 1300 N
D) 1420 N
Correct Answer: B
Solution :
Friction between block A and block B and between block B and surface will oppose the P. \[\therefore \] \[P={{F}_{AB}}+{{F}_{BS}}\] \[={{\mu }_{AB}}{{m}_{A}}g+{{\mu }_{BS}}({{m}_{A}}+{{m}_{B}})g\] \[=0.25\times 100\times 10+\frac{1}{3}(100+200)\times 100\] \[=1250\,N\]You need to login to perform this action.
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