An object, previously at rest, slides $1 m$ $1 m$ down a ramp, with an incline of $\frac{3 π}{8}$ $(3pi)/8$ , and then slides horizontally on the floor for another $2 m$ $2 m$ . If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

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An object, previously at rest, slides $1 m$ $1 m$ down a ramp, with an incline of $\frac{3 π}{8}$ $(3pi)/8$ , and then slides horizontally on the floor for another $2 m$ $2 m$ . If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

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Answer 1 · 2017-07-23T03:34:19

$μ_{k} = 0.388$ $mu_k = 0.388$

Explanation:

WARNING: Long-ish answer

We're asked to find the coefficient of kinetic friction ( $μ_{k}$ $mu_k$ ) of an object sliding down a ramp.

![upload.wikimedia.org]()

We'll split this problem into two parts: the first part is where the object is sliding down the incline, and the second part is where it is sliding across the floor.

Incline:

The expression for the coefficient of kinetic friction $μ_{k}$ $mu_k$ is

$f_{k} = μ_{k} n$ $f_k = mu_kn$

where

$f_{k}$ $f_k$ is the magnitude of the retarding friction force acting as it slides down (denoted $f$ $f$ in the above image)
$n$ $n$ is the magnitude of the normal force exerted by the incline, equal to $m g cos θ$ $mgcostheta$ (denoted $N$ $N$ in the above image)

The expression for the net horizontal force $\sum F_{1 x}$ $sumF_(1x)$ is

$sumF_(1x) = overbrace(mgsintheta)^"gravitational force" - overbrace(f_k)^"kinetic friction force"$

Since $f_k = mu_kn$ , we have

$sumF_(1x) = mgsintheta - mu_kn$

And since the normal force $n = mgcostheta$ ,

$sumF_(1x) = mgsintheta - mu_kmgcostheta$

Using Newton's second law, we can find the expression for the acceleration $a_(1x)$ of the object as it slides down the incline:

$sumF_(1x) = ma_(1x)$

$a_(1x) = (sumF_(1x))/m = (mgsintheta - mu_kmgcostheta)/m = gsintheta - mu_kgcostheta$

What we can now do is apply a kinematics equation for constant acceleration to find the final velocity $v_(1x)$ :

$(v_(1x))^2 = (v_(0x))^2 + 2a_(1x)Deltax$

Here, the initial velocity $v_(0x)$ is $0$ , as we take it to start from rest (no value is given otherwise), and the distance $Deltax$ it travels is the length of the incline ( $1$ $"m"$ ), so we have

$(v_(1x))^2 = (0)^2 + 2(gsintheta - mu_kgcostheta)(1color(white)(l)"m")$

$v_(1x) = sqrt(2(gsintheta - mu_kgcostheta)(1color(white)(l)"m"))$

This velocity is also the initial velocity of the motion along the floor..

Floor:

As the object slides across the floor, the plane is perfectly horizontal, so the normal force $n$ now equals $mg$ . The only horizontal force acting on the object is the kinetic friction force $f_k = mu_kn = mu_kmg$ (which is different than the first one).

The net horizontal force $sumF_(2x)$ is thus

$sumF_(2x) = overbrace(-f_k)^"negative because it opposes motion" = -mu_kmg$

Using Newton's second law again, we can find the floor acceleration $a_(2x)$ :

$a_(2x) = (sumF_(2x))/m = (-mu_kmg)/m = -gmu_k$

We can now use the same constant-acceleration equation as before, but this time the initial velocity

$v_(1x) = sqrt(2(gsintheta - mu_kgcostheta)(1color(white)(l)"m"))color(white)(aaa)$ ( $v_(1x)$ from above)

and the final velocity $v_(2x) = 0$ (it comes to rest).

Plugging in known values, we have

$(0)^2 = (sqrt(2(gsintheta - mu_kgcostheta)(1color(white)(l)"m")))^2 + 2(-gmu_k)(2color(white)(l)"m")$

Rearranging gives

$2(gmu_k)(2color(white)(l)"m") = 2(gsintheta - mu_kgcostheta)(1color(white)(l)"m")$

At this point, we're just solving for $mu_k$ . Dividing both sides by $2g$ :

$mu_k(2color(white)(l)"m") = (sintheta - mu_kcostheta)(1color(white)(l)"m")$

Now, we can divide all terms by $mu_k$ , as well as eliminate the unit $"m"$ :

$2 = (sintheta)/(mu_k) - costheta$

$(sintheta)/(mu_k) = 2+costheta$

Therefore,

$color(red)(mu_k = (sintheta)/(2+costheta)$

Plugging in the angle $theta = (3pi)/8$ , we have

$mu_k = (sin((3pi)/8))/(2+cos((3pi)/8)) = color(blue)(0.388$

Notice how the coefficient doesn't depend on the mass $m$ or gravitational acceleration $g$ if the two surfaces are the same...

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An object, previously at rest, slides $1 m$ $1 m$ down a ramp, with an incline of $\frac{3 π}{8}$ $(3pi)/8$ , and then slides horizontally on the floor for another $2 m$ $2 m$ . If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

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Explanation:

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An object, previously at rest, slides 1 m1m1 m down a ramp, with an incline of (3pi)/8 3π8(3pi)/8 , and then slides horizontally on the floor for another 2 m2m2 m. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

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Explanation:

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An object, previously at rest, slides $1 m$ $1 m$ down a ramp, with an incline of $\frac{3 π}{8}$ $(3pi)/8$ , and then slides horizontally on the floor for another $2 m$ $2 m$ . If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?