Cart friction force

An 80 kg cart moving at constant speed along a horizontal road is subjected to a pulling force of 120 N.
Specify:
a) the magnitude of the frictional force
b) the value of the coefficient of shear friction.

Final Answer:

F₂ =  120 N
k =  0.1529

Step-by-step explanation:

$$\begin{gathered} m=80\text{kg} \\ g=9.81{\text{m/s}}^2 \\ {F}_1=120\text{N} \\ {F}_2=F_1=120=120\text{N} \end{gathered}$$

$$\begin{gathered} G=m\cdot g=80\cdot 9.81=\frac{3924}{5}=784\frac{4}{5}=784.8\text{N} \\ {F}_2=kG \\ k=F_2/G=120/784.8=0.1529 \end{gathered}$$

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