Question

you push a 2kg textbook across a lab table with a force of 8 newtons, it moves with constant velocity. determine the coefficient of kinetic friction between the table and book.

Answer 1

The coefficient of kinetic friction between the table and the book is approximately 0.408.

What is the coefficient of kinetic friction?

The coefficient of kinetic friction can be determined using the formula:

`[ \mu_k = \frac{F_{\text{friction}}}{F_{\text{normal}}} ]`

where

`( \mu_k )` = coefficient of kinetic friction

`F_{\text{friction}}` = force of kinetic friction

`F_{\text{normal}}` = force normal to the surface

Given that the force applied is 8 newtons and the mass of the textbook is 2 kg, the force of kinetic friction can be calculated using the formula:

`[ F_{\text{friction}} = \mu_k * F_{\text{normal}} ]`

The force normal to the surface can be calculated using the formula:

`F_{\text{normal}} = m * g`

where

( m ) = mass of the textbook

( g ) = acceleration due to gravity (approximately 9.81 m/s²

Substituting the values, we get:

`F_{\text{normal}} = 2 , \text{kg} * 9.81 , \text{m/s}^2`

`[ F_{\text{normal}} \approx 19.62 , \text{N} ]`

Now, we can calculate the coefficient of kinetic friction:

`[ \mu_k = \frac{F_{\text{friction}}}{F_{\text{normal}}} = \frac{8 , \text{N}}{19.62 , \text{N}} ]`

= 0.408

Therefore, the coefficient of kinetic friction between the table and the book is approximately 0.408.