question_answer

A small block of mass 1 kg is pushed with a velocity \[2\,\hat{i}\,m{{s}^{-1}}\]over a horizontal plane which itself is moving with a velocity of \[-\,2\,\hat{j}\,m{{s}^{-1}}.\] Coefficient of friction between block and plane is 0.25. Force of friction acting on the block (in newton) nearly is

A) \[17\,(\hat{i}+\hat{j})\]             

B) \[17\,(-\hat{i}-\hat{j})\]

C) \[17\,(\hat{i}-\hat{j})\]

D) \[17\,(-\hat{i}+\hat{j})\]

Correct Answer: C

Solution :

Velocity of block with respect to plane is \[{{v}_{12}}={{v}_{1}}-{{v}_{2}}\]

\[\therefore \]\[{{v}_{12}}=-2\hat{i}-(-2\hat{j})=-2(\hat{i}-\hat{j})\,m{{s}^{-1}}\]

Since, friction opposes this velocity.

So, friction is \[{{f}_{k}}=\mu mg(\hat{n})\]

where, \[\hat{n}\] = unit vector opposite to  \[{{v}_{12}}.\]

\[\therefore \]\[{{f}_{k}}=\frac{0.25\times 1\times 10\times 2(\hat{i}-\hat{j})}{\sqrt{{{2}^{2}}+{{2}^{2}}}}=1.78(\hat{i}-\hat{j})N\]