If an object with a mass of $10 k g$ $10 kg$ is moving on a surface at $4 \frac{m}{s}$ $4 m/s$ and slows to a halt after $6 s$ $6 s$ , what is the friction coefficient of the surface?

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Answer 1 · 2016-08-10T10:21:03

$μ = \frac{1}{15}$ $mu = 1/15$

The initial kinetic energy is lost against the friction losses.

$1/2mv_0^2 = mg mu Delta x$ .

Here $mu$ is the kinetic friction coefficent and $Deltax$ is the path until repos.

Also the movement along the path $Delta x$ is dictated by

$Delta x = v_0 t - 1/2(mu g)t^2$

Solving for $Delta x$ and $mu$ the system

${ (1/2mv_0^2 = mg mu Delta x), (Delta x = v_0 t - 1/2(mu g)t^2) :}$

we obtain

$Delta x = (v_0 t)/2$ and
$mu = v_0/(g t)$

Assuming $g = 10$ we have

$mu = 4/60 = 1/15$

If an object with a mass of 10 kg 10kg10 kg is moving on a surface at 4 m/s4ms4 m/s and slows to a halt after 6 s6s 6 s, what is the friction coefficient of the surface?