Given data:
* The mass of the desk is m = 74 kg.
* The force applied on the desk is F = 270 N.
* The coefficient of friction between the desk and horizontal floor is,
[tex]\mu=0.36\text{ }[/tex]Solution:
The diagrammatic representation of the given system is,
The normal force acting on the desk is,
[tex]N=mg[/tex]where g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} N=74\times9.8 \\ N=725.2\text{ Newton} \end{gathered}[/tex]The frictional force acting on the desk is,
[tex]\begin{gathered} F_r=\mu N \\ F_r=0.36\times725.2 \\ F_r=261.07\text{ N} \end{gathered}[/tex]The net force acting on the desk is,
[tex]\begin{gathered} F_{net}=F-F_r \\ F_{\text{net}}=270-261.07 \\ F_{\text{net}}=8.93\text{ N} \end{gathered}[/tex]Thus, the net force acting on the desk is 8.9 N or approximately 9 N.
According to Newton's second law, the acceleration of the desk is,
[tex]F_{net}=ma[/tex]where a is the acceleration of the desk,
Substituting the known values,
[tex]\begin{gathered} 8.93=74\times a \\ a=\frac{8.93}{74} \\ a=0.12ms^{-2} \end{gathered}[/tex]Thus, the acceleration of the desk is 0.12 meters per second squared.
