Question

A block moves down a smooth inclined plane of inclination ¤. Its velocity on reaching the bottom is v. If it slides down a rough inclined plane of same inclination its velocity on reaching the bottom is v/n where n is a no. greater than 0. The coefficient of friction(mew) is given by.
A) mew= tan¤(1-1/n^2)
B) mew= cot¤(1-1/n^2)
C) mew= tan¤ root 1-1/n^2
D) mew= cot¤ root 1-1/n^2

Answer 1

L is the length of the surface.
On smooth inclined plane there will not be acting any frictional force.
On rough inclined plane there will be frictional force.
We know that,
v²-u²=2aL
For smooth inclined plane,
V²=2aL

V²=2(F/m)L (m is mass)

V²=2[(mgsin∅)/m]L

V²=2(gsin∅)L

v=√(2Lgsin∅)

similarly for rough inclined plane,
Final velocity would be √[2Lg(sin∅-ucos∅)]
Here u is coefficient of friction.

Hence coefficient of friction is
u=tan∅(1-1/n²)