Adock 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.

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

Explanation:

It is given that,

Mass of the crate, m = 24 kg

Force acting on the crate when it is at rest,

When the crate is in motion, force acting on the crate,

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 :

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

Hence, this is the required solution.