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3 years ago

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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 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?

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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