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A box is sliding down an incline tilted at a 11.1° angle above horizontal. The box is initially sliding down the incline at a speed of 1.60 m/s. The coefficient of kinetic friction between the box and the incline is 0.390. How far does the box slide down the incline before coming to rest?

Respuesta :

Answer:s=0.68 m

Explanation:

Given

Inclination [tex]\theta =11.1^{\circ}[/tex]

Speed of block(u)=1.6 m/s

Coefficient of kinetic Friction [tex]\mu _k=0.39[/tex]

deceleration provided by friction=g\sin \theta -\mu _kg\cos \theta [/tex]

Using [tex]v^2-u^2=2as[/tex]

Final velocity v=0

[tex]0-1.6^2=2(g\sin \theta -\mu _kg\cos \theta )s[/tex]

[tex]s=\frac{-1.6^2}{2\cdot (9.8\sin 11.1-0.39\times 9.8\times \cos 11.1)}[/tex]

s=0.68 m

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