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A dock worker loading crates on a ship finds that a 24 kg crate, initially at rest on a horizontal surface, requires a 65 N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 50 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:

[tex]\mu_s=0.276[/tex]

[tex]\mu_k=0.213[/tex]

Explanation:

It is given that,

Mass of the crate, m = 24 kg

Force acting on the crate when it is at rest, [tex]F_r=65\ N[/tex]

When the crate is in motion, force acting on the crate, [tex]F_m=55\ N[/tex]

To find,

The coefficient of static friction between crate and floor

Solution,

When the crate is at rest, the force acting on the crate is given by :

[tex]F_r=\mu_sN[/tex]

[tex]65=\mu_s\times 24\times 9.8[/tex]

[tex]\mu_s=0.276[/tex]

When the crate is in motion, the force acting on the crate is given by :

[tex]F_s=\mu_kN[/tex]

[tex]50=\mu_k\times 24\times 9.8[/tex]

[tex]\mu_k=0.213[/tex]

Hence, this is the required solution.

Hania12

Answer:

0.276

Explanation: Here we have to consider the kinetic friction force acting on the body. If it is going at a constant speed according to newton's 1 st law. There was'not net force induced at the system. There for kinetic friction foce must be equal to the 50N .

But here we have been asked to get cofficient of static friction . There for you have to get static friction force. It should be 65N.

Mass of the object is 24. Then the reaction between surface and the object would be 24*9.8 = 235.2 N

There for using this,

Static friction force = Cofficient of static friction * Reaction

Cofficient of static friction = 65/235.2

= 0.276

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