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Gnoma [55]
3 years ago
11

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 sp

eed 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?
Physics
1 answer:
raketka [301]3 years ago
5 0

Answer:s=0.68 m

Explanation:

Given

Inclination \theta =11.1^{\circ}

Speed of block(u)=1.6 m/s

Coefficient of kinetic Friction \mu _k=0.39

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

Using v^2-u^2=2as

Final velocity v=0

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

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

s=0.68 m

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