Question

A 9.0-kg box of oranges slides from rest down a frictionless incline from a height of 5.0 m. A constant frictional force, introduced at point A, brings the block to rest at point B, 19 m to the right of point A.

What is the coefficient of kinetic friction, k , of the surface from A to B?

Answer 1

The coefficient of kinetic friction is 0.26

Step-by-step explanation:

Given:

Mass of box
`m = 9` kg

Initial height
`h = 5` m

Final height
`h' = 0` m

Distance travel by block
`d = 19` m

For finding coefficient of kinetic friction,

We use conservation laws,

Work done by frictional force is equal to change in energy,

`W_(fr) = \Delta E`

Here only potential energy change so we can write,

`f_(k) d \cos 180 = mg(h') - mg(h)`

Here
`f_(k) = \mu_(k) mg`

`-\mu_(k)* mgd= -mgh`

`\mu_(k) = (h)/(d)`

`\mu _(k) = (5)/(19)`

`\mu _(k) = 0.26`

Therefore, the coefficient of kinetic friction is 0.26