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A dock worker loading crates on a ship finds that a 20 kg crate, initially at rest on a horizontal surface, requires a 89 N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 47 N is required to keep it moving with a constant speed. The acceleration of gravity is 9.8 m/s 2 . Find the coefficient of static friction between crate and floor.

Respuesta :

Answer:0.45

Explanation:

Given

Mass of crate [tex]m=20 kg[/tex]

Force [tex]F=89 N[/tex]

After crate start moving a force of 47 N is required to keep it moving with a constant speed

So we can conclude that 89 N is the force which is required to overcome static friction and after that kinetic friction comes into play which is less than static friction

Static friction[tex]=\mu N[/tex]

where [tex]\mu [/tex]is coefficient of static friction

[tex]N=mg[/tex]

Static Friction[tex]=\mu mg=89[/tex]

[tex]\mu =\frac{89}{20\times 9.8}[/tex]

[tex]\mu =0.45[/tex]

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