Question

HELP PLSSS

A 13.3 kg box sliding across the ground
decelerates at 2.42 m/s2. What is the
coefficient of kinetic friction?
(No unit)
ps( its not 0.242)

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 = (a)/(g)\\\\\mu = (2.42)/(9.8)\\\\\mu = 0.247`

Therefore, the coefficient of kinetic friction is 0.247 .