# An object, previously at rest, slides #6 m# down a ramp, with an incline of #(3pi)/8 #, and then slides horizontally on the floor for another #3 m#. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

##### 2 Answers

#### Answer:

#### Explanation:

The problem may split in two parts:

- The slipping on the ramp.
- The slipping on the horizontal floor.

The reason for that is because the friction force depends on the force which stick the object on the floor and it is different in each section. The linkage between both stretches is the speed of the object just at the end of the ramp.

Let's start with the first stretch:

The force of gravity has two component, the first which parallel to the ramp

#F_x=m*g*sin((3pi)/8)#

and the second one, which is perpendicular to the ramp

#F_y=m*g*cos((3pi)/8)#

The force which moves the ball is the first component of force of gravity minus the friction force,

#F=F_x -F_f = m*g*sin((3pi)/8) - mu * m * g * cos((3pi)/8)#

This will be equivalent to

#F = mg * [sin((3pi)/8 - mu cos((3pi)/8)]#

Thus, the speed at the end of ramp is

#v_1^2 = 0^2 + 2 * a * "6 m"#

#v_1^2 = 2 * F/m * "6 m"#

#v_1^2 = 2 * (cancel(m)g * [sin((3pi)/8 - mu cos((3pi)/8)]))/cancel(m) * "6 m"#

#v_1^2 = 2 * 6 * g[sin((3pi)/8) - mucos((3pi)/8)]" " " "color(red)((1))#

In the 2nd stretch, the only force in action is the friction force,

but now

If you take

#v_2^2 = v_1^2 - 2 * a_h * "3 m"#

#v_1^2 = 2 * a_h * "3 m"#

Here you have

#a_h = F_h/m = (mu * cancel(m) * g)/cancel(m) = mu * g#

which will get you

#v_1^2 = 2 * mu * g * "3 m" = 6 * mu * g" " " "color(red)((2))#

Use equations

#2 * cancel(6 * g) [sin((3pi)/8) - mu cos((3[i)/8)] = mu * cancel(6 * g)#

Rearrange to isolate

#2 sin((3pi)/8) = mu + 2 mu cos((3pi)/8)#

#mu = (2 sin((3pi)/8))/(1 + 2 cos((3pi)/8))#

and since

#sin((3pi)/8) =sqrt( sqrt(2) +2)/2" "# and#" "cos((3pi)/8) =sqrt( (2 - sqrt(2)))/2#

you can say that

#mu = (cancel(2) * sqrt(sqrt(2) + 2)/cancel(2))/(1 + cancel(2) * sqrt(2 - sqrt(2))/cancel(2)) = sqrt(sqrt(2) + 2)/(1 + sqrt(2 - sqrt(2))) ~~ 1.05 #

Since you have

Here length of the ramp (l)= 6m

Angle of inclination of the ramp

Height of the object from horizontal floor,

If mass ofthe body is

The initial gravitational potential energy of the body

The normal reaction acting on the body when it is sliding down the ramp is

and the corresponding frictional force

where

work done against frictional force when sliding down the **ramp**

when the body slides on horizontal force,then normal reaction

work done against frictional force when sliding along **floor**

**Now applying conservation of mechanical energy we can write**

**The initial KE being zero**

Initial PE = total work done against frictional force

**If we consider that the ramp subtends angle**

and the Eq for

**and It is possible**

**Is it OK?**