Question

 A 75 kg box encounters 10 N of friction and slides down a hall with an acceleration of 3.60 m/s2. Find the coefficient of kinetic between the box and the floor. 

Answer 1

Given,

Mass of block m=75kg

Force of friction $f_s=10N$

Acceleration of box a3.60m/$s^2$

Let $\theta$ be the angle of inclination

$\therefore mgsin\theta-f_s=ma$

$\to 75\times9.8\times sin\theta-10=75\times3.60\\\to735sin\theta=10+270\\\to735sin\theta=280\\\to sin\theta=\dfrac{280}{735}\\\to sin\theta=0.3809\\\to\theta=sin^{-1}(0.3809)\\\to \theta=28.38\degree$

As we know frictional force

$f_s=u_s\times R\\\to10=u_k\times mgcos\theta\\\to 10=u_k\times 75\times 9.8\times cos22.38\degree\\\to 10=u_k\times 679.63\\\to u_k=\dfrac{10}{679.63}\\\to u_k=0.0147$

Hence the coefficient of kinetic friction is 0.0147