IIT-JEE 2007 Assertion-Reason Type MCQ on Friction

The following Assertion-Reason Type MCQ which appeared in IIT-JEE 2007 question paper is meant for checking your understanding of the conservative nature of gravitational force and the non-conservative nature of the frictional force:

STATEMENT-1

A block of mass m starts moving on a rough horizontal surface with a velocity v. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of 30º with the horizontal and the same block is made to go up on the surface with the same initial velocity v. The decrease in the mechanical energy in the second situation is smaller than that in the first situation

because

STATEMENT-2

The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination.

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.

(C) Statement-1 is True, Statement-2 is False

(D) Statement-1 is False, Statement-2 is True

In both situations the initial kinetic energy of the block is reduced to zero. In the first situation, the initial K.E. is entirely used up in doing work against friction and is irrecoverably lost since friction is a non-conservative force. In the second situation, part of the initial K.E. is used in doing work against gravitational force and is stored as gravitational potential energy in the block since gravitational force is a conservative force. The decrease in mechanical energy in the second situation is therefore smaller than that in the first situation.

Statement-1 is therefore true.

Statement-2 is false since the coefficient of friction is independent of the inclination and is dependent only on the nature of the two surfaces in contact.

The correct option therefore is (C).

An Interesting Post on Friction

Here is a useful link which will take you to an interesting post on friction:
physicsplus: Friction Moves the Car and Friction Stops the Car

Multiple Choice Questions on Friction:

The following question appeared in the Physics paper of A.I.E.E.E.-2005:

The upper half of an inclined plane with inclination Ф 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 lower half is
(a) tanФ (b) 2tanФ (c) 2cosФ (d) 2sinФ

You have to use the equation of linear motion, v² = u² + 2as for both parts of the motion. For the smooth half of the incline, we have v² = 0 + 2gsinФ×s and for the rough half we have 0 = v² +2(gsinФ - μgcosФ)×s. Combining these two equations, we get
0 = 2gsinФ×s + 2(gsinФ - μgcosФ)×s, from which 2sinФ = μcosФ
Therefore, μ = 2tanФ [Option (b)]
Consider the following MCQ:

A car of mass 1000kg moves on a circular track of radius 20m. If the coefficient of friction is 0.64, then the maximum velocity with which the car can move is
(a) 15m/s (b) 11.2m/s (c) 20m/s (d) 18m/s (e) 22.4m/s
This simple question appeared in the Kerala Engineering Entrance Test paper of 2006. As the frictional force supplies the centripetal force required for the circular motion, we have μmg = mv² /r so that v = √(μrg) = √(0.64×20×9.8) = 11.2m/s

Let us discuss two more questions involving friction:
A coin of mass ‘m’ is placed on a rough horizontal turn table with coefficient of limiting friction ‘μ’. The distance of the coin from the axis of the turn table is ‘r’ and the turn table is turning with angular velocity ‘ω’. The maximum value of ω so as to keep the coin intact on the turn table is
(a) √(μg/r) (b) √(μgr) (c) μgr (d) √(μr/g) (e) μgr²
It is the frictional force that prevents the coin from sliding outwards under the action of the centrifugal force. Therefore we equate the limiting value of the frictional force to the centrifugal force (magnitudes): μmg = mrω² from which ω = √(μg/r). The correct option is therefore (a).
Suppose you are not asked to find the maximum value of ω in the above problem. Instead, you are asked to find the frictional force at the angular velocity ω, as in the modified question below:
A coin of mass ‘m’ is placed on a rough horizontal turn table with coefficient of limiting friction ‘μ’. The distance of the coin from the axis of the turn table is ‘r’ and the turn table is turning with angular velocity ‘ω’. If the coin remains intact on the turn table, the frictional force between the coin and the surface of the turn table is
(a) μmg (b) zero (c) mg (d) mrω² (e) mrω
Nothing is mentioned about the limiting equilibrium of the coin and therefore, the self adjusting frictional force just balances the centrifugal force mrω²
. The correct option therefore is (d).