question_answer

A marble block of mass 2kg lying on ice when given a velocity of 6m/s is stopped by friction in 10s. Then the coefficient of friction is:

A) 0.02                             

B) 0.03   

C) 0.06                             

D) 0.01

Correct Answer: C

Solution :

[c]                           \[{{f}_{r}}=Ma\]             \[\mu R=Ma\]             \[\mu \,Mg=Ma\therefore R=Mg\]             \[\mu =\frac{a}{g}...(1)\]             \[\mu =\frac{6}{10\times 10}=0.06\]             \[v-u=at\]             \[o-u=at[v=o]\]             \[a=\frac{u}{t}=\frac{6}{10}=0.6m/{{s}^{2}}\]             \[put\,value\,in\,(1)\]             \[\mu =\frac{a}{g}=\frac{0.6}{10}=0.06\] \[v-u=at\] \[a=\frac{u}{t}=\frac{6}{10}m/s\] \[v=0\]