Answer:
0.34 m
Explanation:
We are given that
[tex]\theta=11.6^{\circ}[/tex]
Initial speed of box=u=1.1 m/s
Coefficient of friction,[tex]\mu=0.39[/tex]
We have to find the distance slide down by the box before coming to rest.
Friction force=f=[tex]\mu mg=[/tex]
Where [tex]g=9.8 m/s^2[/tex]
Net force=[tex]mgsin\theta-\mu mg[/tex]
[tex]ma=mg(sin\theta-\mu cos\theta)[/tex]
[tex]a=g(sin\theta-\mu cos\theta)[/tex]
Substitute the values
[tex]a=9.8(sin11.6-0.39cos11.6)=9.8(-0.18)=-1.76 m/s^2[/tex]
[tex]v=0[/tex]
[tex]v^2-u^2=2as[/tex]
Substitute the values
[tex]0-(1.1)^2=2(-1.76)s[/tex]
[tex]-3.52s=-(1.1)^2[/tex]
[tex]s=\frac{-(1.1)^2}{-3.52}[/tex]
[tex]s=0.34 m[/tex]
Hence, the box slide down the incline 0.34 m before coming to rest.
