= Sliding acceleration on an inclined plane
{title2=$a=g(\sin\theta-\mu_k\cos\theta)$}
First test rest. With no other tangential <forces>, the required uphill <static friction> is $f_s=mg\sin\theta$. Its maximum is $\mu_sN$, where $\mu_s$ is the <coefficient of static friction>. Thus
$$
mg\sin\theta\le\mu_smg\cos\theta
\quad\Longleftrightarrow\quad
\tan\theta\le\mu_s,\qquad
\theta_c=\arctan\mu_s.
$$
The <tangent function> is increasing on this <angle> range, and its <inverse tangent> determines the last permitted resting <angle> $\theta_c$. At equality, the ideal model still permits limiting rest; a larger <angle> excludes it. Below the threshold, setting $f_s=\mu_sN$ would incorrectly impose the maximum <static friction> even when less is required. <Static friction on an inclined plane> and <inclined-plane measurement of static friction> develop this onset criterion further.
For a block already sliding downhill, uphill <kinetic friction> has magnitude $f_k=\mu_kN$, where $\mu_k$ is the <coefficient of kinetic friction>. Taking downhill as positive, <Newton's second law> becomes
$$
ma=mg\sin\theta-\mu_kmg\cos\theta,
\qquad a=g(\sin\theta-\mu_k\cos\theta).
$$
Every term in the <force> balance scales with <mass>; division by $m>0$ removes it. Doubling <mass> doubles both <gravitational weight> and the contact <forces>, while leaving this <acceleration> unchanged if the coefficients remain unchanged. A mass-independent result is a prediction of these assumptions, rather than a universal property of every ramp.
For an original numerical model, stipulate $m=3\,\mathrm{kg}$, $g=9.8\,\mathrm{m/s^2}$, $\mu_s=0.45$ and $\mu_k=0.30$. The critical <angle> is $\theta_c\simeq24.23^\circ$. At $20^\circ$, the <normal force> is $27.627\,\mathrm N$; rest requires $10.055\,\mathrm N$ of uphill <static friction>, below its $12.432\,\mathrm N$ bound. A resting block can stay there. If it is already sliding downhill, however, the <kinetic friction> is only $8.288\,\mathrm N$, giving downhill <acceleration> $a\simeq0.5891\,\mathrm{m/s^2}$. Specifying the initial state matters.
At $30^\circ$, the downhill component of <gravitational weight> is $14.700\,\mathrm N$, while $N\simeq25.461\,\mathrm N$ and the <static friction> bound is only $11.458\,\mathrm N$. Rest is impossible. During downhill sliding,
$$
f_k\simeq7.638\,\mathrm N,\qquad
F_{\mathrm{net}}\simeq7.062\,\mathrm N,\qquad
a\simeq2.3539\,\mathrm{m/s^2}.
$$
Released from rest on a sufficiently long ramp, this ideal block covers its first $1.00\,\mathrm m$ in $t=\sqrt{2L/a}\simeq0.9218\,\mathrm s$, reaching <speed> $v=\sqrt{2aL}\simeq2.1697\,\mathrm{m/s}$. As an independent <energy> calculation, the lost <gravitational potential energy> is $mgL\sin\theta=14.700\,\mathrm J$ and <kinetic friction> removes $f_kL\simeq7.638\,\mathrm J$ of <mechanical energy>. The remaining $7.062\,\mathrm J$ equals $mv^2/2$ by the <work-energy theorem>.
\Image[/inclined-plane-rest-and-sliding-acceleration.png]
{title=Rest threshold and downhill sliding acceleration for the stipulated inclined-plane model}
{description=Original model predictions with $g=9.8\,\mathrm{m/s^2}$, $\mu_s=0.45$ and $\mu_k=0.30$; no experimental measurements. The shaded range permits rest. The blue curve applies only while downhill <contact slip> persists, including inside that range. The orange curve gives the frictionless <acceleration>. Markers show the $20^\circ$ and $30^\circ$ examples.}
At $\theta=\arctan\mu_k$, downhill sliding has zero <acceleration> and therefore constant <velocity>. Below that <angle>, a downhill-moving block slows; when it stops, test the <static friction> bound again. For uphill sliding, <kinetic friction> points downhill, so the signed downhill <acceleration> instead is $g(\sin\theta+\mu_k\cos\theta)$. The sign follows <contact slip>, and the downhill formula cannot be used unchanged after a reversal.
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