Question

HELP PLSSS

Answer 1

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 :

$f = ma$    ....1)

Also, force of friction is given by :

$f = \mu mg$  ....2)

Equating equation 1) and 2), we get :

$\mu mg = ma\\\\\mu = \dfrac{a}{g}\\\\\mu = \dfrac{2.42}{9.8}\\\\\mu = 0.247$

Therefore, the coefficient of kinetic friction is 0.247 .