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.

Answers

Answer 1

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

Answer 2

Answer:

Answer:

$\mu_s=0.276$

$\mu_k=0.213$

Explanation:

It is given that,

Mass of the crate, m = 24 kg

Force acting on the crate when it is at rest, $F_r=65\ N$

When the crate is in motion, force acting on the crate, $F_m=55\ N$

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 :

$F_r=\mu_sN$

$65=\mu_s* 24* 9.8$

$\mu_s=0.276$

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

$F_s=\mu_kN$

$50=\mu_k* 24* 9.8$

$\mu_k=0.213$

Hence, this is the required solution.

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