Given :
A 13.3 kg box sliding across the ground decelerates at 2.42 m/s².
To Find :
The coefficient of kinetic friction.
Solution :
Frictional force applied to the box is :
[tex]f = ma[/tex] ....1)
Also, force of friction is given by :
[tex]f = \mu mg[/tex] ....2)
Equating equation 1) and 2), we get :
[tex]\mu mg = ma\\\\\mu = \dfrac{a}{g}\\\\\mu = \dfrac{2.42}{9.8}\\\\\mu = 0.247[/tex]
Therefore, the coefficient of kinetic friction is 0.247 .
