The given problem can be exemplified in the following free-body diagram:
Now we sum the forces in the horizontal direction:
[tex]F-F_f=ma_x[/tex]Where:
[tex]\begin{gathered} F=\text{ Horizontal force} \\ F_f=\text{ Friction force} \\ m=\text{ mass} \\ a=\text{ acceleration} \end{gathered}[/tex]Since the movement is required to be at a constant speed the acceleration must be zero, therefore:
[tex]F-F_f=0_{}[/tex]Now we solve for the force "F":
[tex]F=F_f[/tex]Now we use the following definition of friction:
[tex]F_f=\mu_kN[/tex]Where:
[tex]\begin{gathered} \mu_k=\text{ coefficient of kinetic friction} \\ N=\text{ normal force} \end{gathered}[/tex]The kinetic friction coefficient is used due to the fact that we are considering the body as it is already in movement.
Replacing in the sum of horizontal forces:
[tex]F=\mu_kN[/tex]Now, to determine the value of the normal force we will add the forces in the vertical direction:
[tex]N-mg=0[/tex]The vertical forces add up to zero because there is no vertical movement. Now we solve for the normal force "F":
[tex]N=mg[/tex]Now we replace in the sum of horizontal forces:
[tex]F=mg\mu_k[/tex]Now we plug in the values:
[tex]F=(20\operatorname{kg})(10\frac{m}{s^2})(0.6)[/tex]Solving the operations we get:
[tex]F=120N[/tex]Therefore, a force of 120 Newtons is required.
