Answer:
Explanation:
Given
at [tex]\theta =30^{\circ}[/tex]
Box slides down [tex]2.5 m\ in\ t=4 s[/tex]
as box just starts to slide at [tex]\theta =30^{\circ}[/tex]
friction force will just balance the sin component of weight
thus
[tex]mg\sin \theta =f_r[/tex]
and [tex]f_r=\mu N[/tex]
where [tex]\mu [/tex]is coefficient of static friction
[tex]N=mg\cos \theta [/tex]
thus [tex]\mu =\tan \theta [/tex]
[tex]\mu =0.577[/tex]
(b)block travels 2.5 m in 4 s
using [tex]s=ut+\frac{at^2}{2}[/tex]
[tex]2=0+\frac{a\times 16}{2}[/tex]
[tex]a=0.25 m/s^2[/tex]
i.e. [tex]a=\mu _kg[/tex]
[tex]\mu_k=0.025[/tex]
