Respuesta :
Given data:
* The acceleration of the car is,
[tex]a=5ms^{-2}[/tex]* The mass of the car is,
[tex]m=1500\text{ kg}[/tex]* The force acting on the car in the upward direction is,
[tex]F_1=600\text{ N}[/tex]* The force acting on the car in the downward direction is,
[tex]F_2=1000\text{ N}[/tex]* The coefficient of kinetic friction is,
[tex]\mu_k=1[/tex]Solution:
The weight of the car is,
[tex]\begin{gathered} w=mg \\ w=1500\times9.8 \\ w=14700\text{ N} \end{gathered}[/tex]The normal force acting on the car is,
[tex]\begin{gathered} F_N=w+F_2-F_1 \\ F_N=14700+1000-600 \\ F_N=15100\text{ N} \end{gathered}[/tex]The frictional force acting on the car is,
[tex]\begin{gathered} F_k=\mu_kF_N \\ F_k=1\times15100 \\ F_k=15100\text{ N} \end{gathered}[/tex]According to newton's second law, the force acting on the car is,
[tex]\begin{gathered} F=ma \\ F=1500\times5.0 \\ F=7500\text{ N} \end{gathered}[/tex]The net force acting on the car in terms of the applied force and frictional force is,
[tex]\begin{gathered} F=F_a-F_k \\ F_a=F+F_k \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} F_a=7500+15100 \\ F_a=22600\text{ N} \end{gathered}[/tex]Thus, the driving force required to maintain the motion of the car is 22600 N.
