Question

Consider a box sliding down an incline of `45˚` . It takes the block `36.4` cm to increase it's speed from rest to 2 m/s. What is the coefficient of kinetic friction?

Answer 1

The coefficient of friction is `0.21`

Explanation:

Try drawing a free body diagram--it helps!

-The force of gravity acts straight down.
-The normal force acts upwards, in a perpendicular direction from the plane
-The force of friction acts up the incline.

The first step in this problem will be finding the expression for net force. The box will be sliding down due to the horizontal component of velocity, or `mgsintheta`. The force of friction, `mu mg costheta`, will be slowing it down.

Thus:

`F_"net" = mgsintheta - mu mg costheta`

`ma = m(gsintheta - mugcostheta)`

`a = gsintheta - mu g costheta`

We are looking for `mu`, the coefficient of friction. Thus:

`mugcostheta = gsintheta - a`

`mu = (gsintheta - a)/(gcostheta)`

`mu = tan theta - a/(gcostheta)`

So all we have to do is find the acceleration involved. We are given that the velocity increases to 2 `m/s` from `0 m/s` in `36.4 cm`. Thus, using `v^2 = (v_0)^2 + 2a(x - x_0)`

But before we fill in our parameters, we must notice that `36.4 cm = 0.364 m`

`2^2 = 0^2 + 2a(0.364)`

`4 = 0.728a`

`a = 4/(0.728)`

Now we can substitute:

`mu = tan(45˚) - (4/0.728)/(9.8cos(45˚))`

`mu = 0. 21`

Hopefully this helps!