A plank with a box on it at one end is gradua | vedclass

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(C) Let $\mu_s$ and $\mu_k$ be the coefficients of static and kinetic friction between the box and the plank,respectively.When the angle of inclinatio

Vedclass · Accepted Answer

$0.6$ and $0.5$

Vedclass · Answer

(C) Let $\mu_s$ and $\mu_k$ be the coefficients of static and kinetic friction between the box and the plank,respectively.When the angle of inclination $	heta$ reaches $30^o$,the block just starts to slide. Therefore,$\mu_s = 	an	heta = 	an 30^o = \frac{1}{\sqrt{3}} \approx 0.577 \approx 0.6$.If $a$ is the acceleration produced in the block,then the equation of motion is:$ma = mg\sin	heta - f_k$$ma = mg\sin	heta - \mu_k N$Since $N = mg\cos	heta$,we have:$a = g(\sin	heta - \mu_k\cos	heta)$Given $g = 10\, m/s^2$,$	heta = 30^o$,$s = 4.0\, m$,and $t = 4.0\, s$,we use the kinematic equation $s = ut + \frac{1}{2}at^2$ (with $u = 0$):$4.0 = 0 + \frac{1}{2} a (4.0)^2$$4.0 = 8a \implies a = 0.5\, m/s^2$.Substituting $a$ into the acceleration equation:$0.5 = 10(\sin 30^o - \mu_k \cos 30^o)$$0.5 = 10(0.5 - \mu_k \frac{\sqrt{3}}{2})$$0.05 = 0.5 - \mu_k \frac{\sqrt{3}}{2}$$\mu_k \frac{\sqrt{3}}{2} = 0.45$$\mu_k = \frac{0.9}{\sqrt{3}} = \frac{0.9}{1.732} \approx 0.519 \approx 0.5$.Thus,$\mu_s \approx 0.6$ and $\mu_k \approx 0.5$.

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