A body slides down a plane 45‘ incline in twice the time it takes to slide down a 45‘ smooth plane of  the same length. What is the coefficient of friction between the body  and the incline?

‘ indicates Degree.

Dear Student,

It is given that along a plane inclined at 450, a block slides and takes double the amount of time than it takes to slide along the same block if it is smooth i.e. the force of friction is not there hence the coefficient of friction will be 0 for the smooth incline. As we can see from the adjoining diagram,

mg is the force or weight of the block pulling it downwards.

Hence mg cosӨ is the component which exactly balances the Normal component of force N.

mg sinӨ is the component of force pushing the block along the plane.

The friction component, f , acts opposite to mg sinӨ. 

Now let us consider the two different cases.

For a smooth plane,

f = 0 (as we assume friction to be 0 in this case)

N and mg cosӨ balance and cancel each other. 

Let the acceleration of the block in this case be ‘a’. hence force will be 

F = ma,

This force will be equal to the net force hence,

ma = mg sinӨ- f,

f = 0 hence,

ma = mg sinӨ

a = g sinӨ. ………………………………..(1)

Now for the next case, there is friction,

Friction (f) = coefficient of friction * component of force

Hence it is μmg cosӨ

let the acceleration be a',

hence the equation for force will become,

ma' = mg sinӨ – μmg cosӨ

a' = g sinӨ - μgcosӨ ………………….(2)

now, let us briefly use the equations of motions to find the relation between a and a',

from the 2nd equation of motion we know,

s = ut + ½ at²

where s is the displacement, u is the initial velocity (which is 0 for both the cases) and t is the time taken to cover ‘s’. The block is initially at rest before it starts sliding hence u = 0.

s = ½ at² for case 1

time taken to cover the rough incline is twice the time taken to cover the smooth incline hence equation of motion becomes,

s = ½ a'(2t)²

s = ½ a'*4t²

equating the above two equations, we get,

a/4 = a'

substituting the value of a/in the eqn no : 2

a/4 = g sinӨ –μg cosӨ

(g sinӨ)/4 = g sinӨ – μg cosӨ…. ……….(from eqn 1)

(g sinӨ)/4 – g sinӨ = -μg cosӨ

-3/4(g sinӨ) = -μg cosӨ

Now by comparing the corresponding quantities we get,

sinӨ = cosӨ for Ө = 45^o, 

-3/4 = - μ

¾ = μ which is the required value for coefficient of friction.