Question

A 2.4 kg block slides freely across a rough horizontal surface such that the block slows down with an acceleration of -0.8 m/s^2. What is the coefficient of kinetic friction between the block and the surface? (Use g= 10 m/s^2) *

Answer 1

Given:

Mass of block = 2.4 kg

Acceleration = -0.8 m/s^2

Let's ifnd the coefficient of kinetic friction,

To find the coefficient of kinteic friction, apply the formula:

`\mu=(F)/(N)`

Where:

F is frictional force

N is the normal force.

Hence, we have:

`\begin{gathered} \mu=(ma)/(mg) \\ \\ \mu=(2.4*(0.8))/(2.4*10) \\ \\ \mu=(1.92)/(24) \\ \\ \mu=0.08 \end{gathered}`

Therefore, the coefficient of friction between the block and the surface is 0.08

ANSWER:

0.08