Question

The upper half of an inclined plane with inclination $\varphi$ is perfectly smooth, while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom, if the coefficient of friction for the lower half is given by

• A. $2\sin \varphi$
• B. $2\cos \varphi$
• C. $2\tan \varphi$
• D. $\tan \varphi$

Answer 1

The correct option is C $2\tan \varphi$

K.E of a block at point P is given as

K.E $=mg-\frac{l}{2}\sin \varphi$

At point C , K.E of the block will be zero , therefore K.E is lost in work done by friction.

$W=\mu N\frac{l}{2}=\mu mg\frac{l}{2}\cos \varphi$

By energy conservation at point B and C;

At B,

total energy $=K.E+P.E=K.E+mgh=K.E+mg\left(\frac{l}{2}\sin \varphi \right)$$=mg\left(\frac{l}{2}\sin \varphi \right)$$+mg\left(\frac{l}{2}\sin \varphi \right)$$=mgl\sin \varphi$

Now, $W$ = total energy at B

Since potential energy at Cis zero.

$\mu \left(\frac{1}{2}\right)mgl\cos \varphi =mgl\sin \varphi$

$\mu =\frac{2\sin \varphi }{\cos \varphi }=2\tan \varphi$