Question

A body is pushed up on a rough inclined plane making an angle of 30° to the horizontal. If its time of
ascent on the plane is half the time of descent, find coefficient of friction​

Answer 1

Given info : A body is pushed up on a rough inclined plane making an angle of 30° to the horizontal. If its time of

ascent on the plane is half the time of descent.

To find : the coefficient of friction.

Solution : first find acceleration of body in both cases.

Case 1 : when body ascends along plane.

Friction opposes relative motion so, it acts just opposite to its motion.

then, net acceleration acting on body, a = (gsin30° + μgcos30°)

[ I hope you understand how did I get it ? okay, friction force and weight both acting in same direction here weight along plane is mgsin30° and frictional force is μmgcos30° so net acceleration = net force/mass = (mgsin30° + μmgcos30°)/m]

Case 2 : when body descends

Friction force just acts upward on the body along plane. So, net force must be F = (mgsin30° - μmgcos30°)

So, net acceleration, a' = (gsin30° - μgcos30°)

Now if length of inclined plane is S, then S = 1/2 at² ⇒t = √(2S/a)

And S = 1/2 a't'² ⇒t' = √(2S/a')

Where t and t' are time of ascent and time of descent respectively.

A/c to question,

t = t'/2

⇒√(2S/a) = √(2S/a')/2

⇒8S/a = 2S/a'

⇒4a' = a

⇒4(gsin30° - μgcos30°) = gsin30° + μgcos30°

⇒3gsin30° = 5μgcos30°

⇒μ = (3/5)tan30° = 3/5√3

therefore the coefficient of friction is 3/5√3

Answer 2

Given info : A body is pushed up on a rough inclined plane making an angle of 30° to the horizontal. If its time of

ascent on the plane is half the time of descent.

To find : the coefficient of friction.

Solution : first find acceleration of body in both cases.

Case 1 : when body ascends along plane.

Friction opposes relative motion so, it acts just opposite to its motion.

then, net acceleration acting on body, a = (gsin30° + μgcos30°)

[ I hope you understand how did I get it ? okay, friction force and weight both acting in same direction here weight along plane is mgsin30° and frictional force is μmgcos30° so net acceleration = net force/mass = (mgsin30° + μmgcos30°)/m]

Case 2 : when body descends

Friction force just acts upward on the body along plane. So, net force must be F = (mgsin30° - μmgcos30°)

So, net acceleration, a' = (gsin30° - μgcos30°)

Now if length of inclined plane is S, then S = 1/2 at² ⇒t = √(2S/a)

And S = 1/2 a't'² ⇒t' = √(2S/a')

Where t and t' are time of ascent and time of descent respectively.

A/c to question,

t = t'/2

⇒√(2S/a) = √(2S/a')/2

⇒8S/a = 2S/a'

⇒4a' = a

⇒4(gsin30° - μgcos30°) = gsin30° + μgcos30°

⇒3gsin30° = 5μgcos30°

⇒μ = (3/5)tan30° = 3/5√3

therefore the coefficient of friction is 3/5√3