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

Answer:

The coefficient of kinetic friction is 0.26

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 $f_{k} = \mu_{k} mg$

$-\mu_{k}\times mgd= -mgh$

  $\mu_{k} = \frac{h}{d}$

  $\mu _{k} = \frac{5}{19}$

  $\mu _{k} = 0.26$

Therefore, the coefficient of kinetic friction is 0.26