A) \[\mu =\frac{1}{\tan \theta }\]
B) \[\mu =\frac{2}{\tan \theta }\]
C) \[\mu =2\tan \theta \]
D) \[\mu =\tan \theta \]
Correct Answer: C
Solution :
Using work-energy theorem \[{{W}_{total}}=\Delta K\] \[{{W}_{gravity}}+{{W}_{friction}}=(0-0)=0\] \[(mg\sin \theta .L)+\left( -\mu mg\cos \theta \times \frac{L}{2} \right)=0\] \[\Rightarrow \]\[\mu =\frac{2\sin \theta }{\cos \theta }=2\tan \theta \]You need to login to perform this action.
You will be redirected in
3 sec