A) 11.9 N
B) 25 N
C) 50 N
D) 22.9 N
Correct Answer: A
Solution :
The frictional force acting on the block is \[{{f}_{s}}={{\mu }_{s}}R\] But, \[R=mg\cos {{30}^{o}}\] \[\therefore \] \[{{f}_{s}}=\mu mg\cos {{30}^{o}}\] \[\Rightarrow \] \[{{f}_{s}}=0.7\times 2\times 9.8\times 0.866\] \[\Rightarrow \] \[{{f}_{s}}=11.9\,N\]You need to login to perform this action.
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