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,\[R=mg\cos {{30}^{o}}\] \[\therefore \] \[{{f}_{s}}=\mu mg\cos {{30}^{o}}\] \[\Rightarrow \] \[{{f}_{s}}=0.7\times 9.8\times 0.866\] \[\Rightarrow \] \[{{f}_{s}}=11.9\,\,N\]You need to login to perform this action.
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