Question #b79c6

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Question #b79c6

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Answer 1 · 2016-11-27T00:25:42

Let
$m \to mass of the block$ $m->"mass of the block"$
$θ \to angle of inclination of the inclined plane = {29.7}^{\circ}$ $theta->"angle of inclination of the inclined plane "=29.7^@$
$g \to acceleration due to gravity$ $g->"acceleration due to gravity"$
$μ_{k} \to coefficient of kinetic friction$ $mu_k->"coefficient of kinetic friction "$

Since in this case the block moves downward with constant speed the net force acting on it zero,
Here downward gravitational pull along the plane $m g sin θ$ $color(blue)(mgsintheta)$ acting on the block is just balanced by the upward frictional force $μ_{k} m g cos θ$ $color(red)(mu_kmgcostheta)$ along the plane

So $m g sin θ = μ_{k} m g cos θ$ $mgsintheta=mu_kmgcostheta$

$\Rightarrow μ_{k} = tan θ = {tan 29.7}^{\circ} = 0.57$ $=>mu_k=tantheta=tan29.7^@=0.57$

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