Properties of Friction — Question 7

Question 7

A rough patch can reveal where a sliding body stops

Take $g=10$ m/s$^2$. Use ideal dry friction, $|f_s|\le\mu_sN$ and $|f_k|=\mu_kN$. A small block enters a 2 m long horizontal patch at 4 m/s. That patch has $\mu_k=0.20$; beyond it a second patch has $\mu_k=0.40$. No horizontal applied force acts. Ignore block length.

1. Determine whether it reaches the second patch and find its entry speed.

2. Find the stopping distance measured from the start of the first patch.

3. Find total stopping time and explain why using an average coefficient over the distance is unnecessary.

Question 7 - Solution

A rough patch can reveal where a sliding body stops

(a) Check speed squared at the boundary.

In patch 1, $a_1=-\mu_{k1}g=-2$ m/s$^2$. At its far boundary, $$v_b^2=v_0^2+2a_1L_1=16+2(-2)(2)=8>0.$$ The block reaches patch 2 with $\boxed{v_b=2\sqrt2\ \mathrm{m/s}}$.

(b) Stop under the second acceleration.

Patch 2 supplies $a_2=-4$ m/s$^2$. Its stopping segment satisfies $$0=v_b^2+2a_2L_2\Longrightarrow L_2=\frac8{8}=1\ \mathrm m.$$ $$\boxed{L_{\mathrm{stop}}=L_1+L_2=2+1=3\ \mathrm m}.$$

(c) Add the two durations.

Use each phase’s acceleration and endpoint speeds: $$t_1=\frac{2\sqrt2-4}{-2}=2-\sqrt2\ \mathrm s,\qquad t_2=\frac{0-2\sqrt2}{-4}=\frac{\sqrt2}{2}\ \mathrm s.$$ $$\boxed{t_{\mathrm{stop}}=2-\frac{\sqrt2}{2}=1.2929\ \mathrm s}.$$ No averaged coefficient is needed; the boundary speed connects the two exact constant-acceleration solutions.