question_answer

A force of 49 N is just able to move a block of wood weighing 10 kg on a rough horizontal surface. Its coefficient of friction is

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,\[F=49\,\,N,\,\,m=10\,\,kg,\,\,g=10\,m/{{s}^{2}}\] \[\therefore \]  \[\mu =\frac{49}{10\times 10}=0.49\approx 0.5\]