Question

A dock worker loading crates on a ship finds that a 27 kg crate, initially at rest on a horizontal surface, requires a 80 N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 56 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.

Answer 1

Answer:0.302

Explanation:

Given

mass of crate m=27 kg

Force required to set crate in motion is 80 N

Once the crate is set  in motion 56 N is require to move it with constant velocity

i.e. 80 N is the amount of force needed to just overcome static friction and 56 is the kinetic friction force

thus  

$f_s(static\ friction)=\mu \cdot N$

where $\mu$is the coefficient of static friction and N is Normal reaction

$N=mg$

$f_s=\mu mg$

$\mu mg=80$

$\mu =\frac{80}{27\times 9.8}$

$\mu =0.302$