A) \[1\]
B) 0.7
C) 0.5
D) zero
Correct Answer: C
Solution :
When the applied force \[F\] is increased the force of static friction \[({{f}_{s}})\] also increases, but after a certain limit \[{{f}_{s}}\] cannot increase any more. At this moment block is just to move \[\therefore \] \[F={{f}_{s}}\] and \[{{f}_{s}}=\mu R\] where \[R\] is the reaction of the surface on the block. \[\therefore \] \[F=\mu R\] \[\Rightarrow \] \[\mu =\frac{F}{R}=\frac{F}{mg}\] Given,\[F=49\,\,N,\,\,m=10\,\,kg,\,\,g=10\,m/{{s}^{2}}\] \[\therefore \] \[\mu =\frac{49}{10\times 10}=0.49\approx 0.5\]You need to login to perform this action.
You will be redirected in
3 sec