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) *

Respuesta :

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:

[tex]\mu=\frac{F}{N}[/tex]

Where:

F is frictional force

N is the normal force.

Hence, we have:

[tex]\begin{gathered} \mu=\frac{ma}{mg} \\ \\ \mu=\frac{2.4\times(0.8)}{2.4\times10} \\ \\ \mu=\frac{1.92}{24} \\ \\ \mu=0.08 \end{gathered}[/tex]

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

ANSWER:

0.08

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