Question

A doc worker loading crates on a ship finds that a 18 kg crate, initially at rest on a horizontal surface requires a 80N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 46 N is required to keep it moving with a constant speed. The acceleration of gravity is 9.8m/s^2. Find the coefficient of friction between crate and floor

Answer 1

Answer:$c_{s}=0.45$,$c_{k}=0.26$

Explanation:

Let the coefficient of static friction be $c_{s}$

Let the coefficient of kinetic friction be $c_{k}$

Let the normal reaction force be $N$

$N=mg$ where $m$ is the mass of crate and $g$ is the universal gravitational constant.

Given,

$m=18Kg\\g=9.8ms^{-2}$

$N=18\times 9.8=176.4N$

It is given that $80N$ is required to overcome static friction,

So,$80=c_{s}\times 176.4$

$c_{s}=0.45$

It is given that $46N$ is required to overcome kinetic friction,

So,$46=c_{k}\times 176.4$

$c_{k}=0.26$