Question

A boy pushes a 74kg desk across the floor using a force of 270N. The horizontal floor has a coefficient of friction of 0.36. What is the Net force on the desk? What is the acceleration of desk?

Answer 1

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,

`\mu=0.36\text{ }`

Solution:

The diagrammatic representation of the given system is,

The normal force acting on the desk is,

`N=mg`

where g is the acceleration due to gravity,

Substituting the known values,

`\begin{gathered} N=74*9.8 \\ N=725.2\text{ Newton} \end{gathered}`

The frictional force acting on the desk is,

`\begin{gathered} F_r=\mu N \\ F_r=0.36*725.2 \\ F_r=261.07\text{ N} \end{gathered}`

The net force acting on the desk is,

`\begin{gathered} F_(net)=F-F_r \\ F_{\text{net}}=270-261.07 \\ F_{\text{net}}=8.93\text{ N} \end{gathered}`

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,

`F_(net)=ma`

where a is the acceleration of the desk,

Substituting the known values,

`\begin{gathered} 8.93=74* a \\ a=(8.93)/(74) \\ a=0.12ms^(-2) \end{gathered}`

Thus, the acceleration of the desk is 0.12 meters per second squared.