Question

a rough block slides down the entire length of a smooth inclined plane (angle theta). A similar but smooth block would take half the time to slide the entire length. Find the coefficient of kinetic frictions between the rough block and the incline in terms of the angle theta

Answer 1

The value is
`\mu_k  =  (3)/(4) *  tan(\theta )`

Step-by-step explanation:

Generally the net force acting on the smooth block is mathematically represented as

`F  =  mgsin (\theta )`

Here this force also represented as

`F =  ma`

So

`ma  =  mgsin (\theta )`

=>
`a =  g sin(\theta )`

Generally from kinematic equation

`s =  ut  + (1)/(2) a t_1^2`

before the sliding the block was at rest so u = 0 m/s

`s =  (1)/(2) a t_1^2`

=>
`s =  (1)/(2) (g sin(\theta )) t^2`

=>
`t_1 =  \sqrt{(2s)/(gsin(\theta )) }`

Generally the net force acting on the rough block is mathematically represented as

`F  = mgsin(\theta )  - F_f`

Here
`F_f` is the frictional force acting on the block which is mathematically represented as

`F_f  =  \mu_k  *  mg *  cos (\theta)`

So

`F  = mgsin(\theta )  -  \mu_k  *  mg *  cos (\theta)`

generally this net force is mathematically represented as

`F =  ma`

=>
`ma  = mgsin(\theta )  -  \mu_k  *  mg *  cos (\theta)`

=>
`a = gsin(\theta ) - \mu_k * cos (\theta )`

Generally from kinematic equation

`s =  ut  + (1)/(2) a t_2^2`

before the sliding the block was at rest so u = 0 m/s

`s =  (1)/(2) a t_2^2`

=>
`s =  (1)/(2) [gsin(\theta ) - \mu_k * cos (\theta )} t^2`

= >
`t_2  =  \sqrt{(2s)/(gsin(\theta ) - \mu_k cos(\theta )) }`

From the question we are told that

`t_2  =  2 t_1`

`\sqrt{(2s)/(gsin(\theta ) - \mu_k cos(\theta )) } = 2 * \sqrt{(2s)/(gsin(\theta )) }`

Squaring both sides

`(1)/(sin(\theta ) -  \mu_k cos(\theta ))  =  (2)/(sin(\theta))`

=>
`sin(\theta) =  4sin(\theta ) -  4\mu_k cos(\theta )`

=>
`4\mu_k cos(\theta ) =  3sin(\theta )`

=>
`\mu_k  =  (3)/(4) * (sin(\theta ))/( cos(\theta ))`

=>
`\mu_k  =  (3)/(4) *  tan(\theta )`