Question

A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches 30° , the box starts to slip and slides 4.0 m down the plank in 4.0 s. The coefficients of static and kinetic friction between the box and the plank will be, respectively?​

Answer 1

$\huge\sf\pink{Solution}$

Static coefficient of friction is μs = tan30° = 0.577 ≈ 0.6

For kinetic friction, ma = mg sin30 - f = mg sin30 μk

mg cos30

a = g sin30 - μk

g cos30

and also using S = $ut+1/2at^2$ ,

⇒ $</strong><strong>4</strong><strong> </strong><strong>=</strong><strong> </strong><strong>0</strong><strong>+</strong><strong>(</strong><strong>1</strong><strong>/</strong><strong>2</strong><strong>)</strong><strong>a</strong><strong>(</strong><strong>4</strong><strong>)</strong><strong>^</strong><strong>2</strong><strong>$

or $a = 0.5m/s^2$

Now from (1) we get, 0.5=10(1/2)−μk (10) $\frac{ \sqrt{3}}{2}$

or μk $\frac{4.5}{5 \sqrt{3}}$ = 0.5

Answer 2

hope it's helpful!!! ☺☺