An object, previously at rest, slides $9 m$ $9 m$ down a ramp, with an incline of $\frac{π}{3}$ $(pi)/3$ , and then slides horizontally on the floor for another $12 m$ $12 m$ . 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 $9 m$ $9 m$ down a ramp, with an incline of $\frac{π}{3}$ $(pi)/3$ , and then slides horizontally on the floor for another $12 m$ $12 m$ . 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 · 2017-05-10T13:08:13

The frictional coefficient of the surface is given by $μ = \times$ $\mu = xx$

Frictional coefficients have no units.

Explanation:

We are not told the mass of the object, but we can just call it $m$ $m$ , and I have a feeling it will cancel out anyway.

Two forces due to the weight force of gravity on the object's mass are acting while the object is on the slope: a force parallel to the slope, $F_{//} = m g sin θ$ $F_"//" = mgsin\theta$ , and a force perpendicular to the slope, $F_{⊢} = m g cos θ$ $F_|-- = mgcos\theta$ .

A third force is also acting, the frictional force, given by $F_{frict} = μ F_{⊢}$ $F_"frict" = \muF_|--$ where $μ$ $\mu$ is the coefficient of friction. The frictional force acts in the opposite direction to $F_{//}$ $F_"//"$ .

When the object is moving across the flat floor, only the frictional force is acting. In this case, $θ = 0^{o}$ $\theta = 0^o$ and $cos θ = 1$ $cos \theta = 1$ , so the frictional force (it will be different from that when the object on the ramp although the frictional coefficient is the same) is given by $F_{fric} = μ m g$ $F_"fric" = \mumg$ .

The object begins and ends its journey at rest. In the mean time, it accelerates while on the inclined plane and decelerates while on the floor.

Its deceleration in the floor will be given by $f = \frac{F_{fric}}{m} = \frac{μ m g}{m} = μ g$ $f=F_"fric"/m = (\mumg)/m = \mug$

As I suspected, the mass of the object cancels out. We can find its velocity, $v$ $v$ at the bottom of the ramp, at least in terms of $μ$ $\mu$ , by solving $v^{2} = u^{2} + 2 a s$ $v^2 = u^2 + 2as$ (some people write the last term as ' $2 a d$ $2ad$ ') where $v$ $v$ , the final velocity on the floor, is $0$ $0$ $m s^{- 1}$ $ms^-1$ and $u$ $u$ , the initial velocity on the floor, is what we want to find. I won't go through all of the solution here for space reasons, but substituting in the $12$ $12$ $m$ $m$ traveled I get $u =$ $u =$ $m s^{- 1}$ $ms^-1$ .

{sigh - trying hard but struggling on this one and sleepy - will return to wrap it up tomorrow}

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An object, previously at rest, slides $9 m$ $9 m$ down a ramp, with an incline of $\frac{π}{3}$ $(pi)/3$ , and then slides horizontally on the floor for another $12 m$ $12 m$ . If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

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