Laws Of Motion Question 386
Question: The time taken by a body in sliding down a rough inclined plane of angle of inclination $ 45{}^\circ $ is n times the time taken by the same body in slipping down a similar frictionless plane. The coefficient of dynamic friction between the body and the plane will be
Options:
A) $ 1/(1-n^{2}) $
B) $ 1-(1/n^{2}) $
C) Ö$ { 1-(1/n^{2}) } $
D) Ö$ { 1/(1-n^{2}) } $
Show Answer
Answer:
Correct Answer: B
Solution:
Acceleration when there is no friction is $ a=g\sin \theta $ and acceleration when friction is there is
$ a’=(g\sin \theta -\mu g\cos \theta ) $
We know that $ s=ut+\frac{1}{2}at^{2}, $ under condition of $ u=0, $
gives,$ t=\sqrt{( 2s/a )} $ .
Therefore $ t’t=\sqrt{( a/a’ )} $ .
For our case we get
$ n=\sqrt{[g\sin \theta /(g\sin \theta -\mu g\cos \theta )]} $
$ \Rightarrow n^{2}=g\sin \theta /(g\sin \theta -\mu g\cos \theta )=1/(1-\mu \cot \theta ) $ but $ \theta =45^{o} $
$ \Rightarrow \cot \theta =1\therefore n^{2}=1/(1-\mu ) $ or$ \mu =1-(1/n^{2}) $
BETA
