Answer:
a. [tex]\mathbf{F_{f_2} = 2 F_{f1}}[/tex], [tex]\mathbf { since \ the \ force \ o f \ friction \ is \ represented \ as \ the \ slope \ for \ each \ of \ the \ two \ curves.}[/tex]
Explanation:
From the information given;
Using the work-energy theorem
ΔKE = W = [tex]\mathbf{ F_f \times r}[/tex]
K = [tex]\mathbf{ F_f \times r}[/tex]
∴
[tex]\dfrac{K_1}{K_2} = \dfrac{F_{f1}}{F_{f2}} (\dfrac{r_1}{r_2})[/tex]
Since [tex]K_1 = K_2[/tex] and r_1 = 4, and r_2 = 2 (from the missing diagram which is attached below)
Then;
[tex]1 = \dfrac{F_{f1}}{F_{f2}} (\dfrac{4 \ m}{2 \ m})[/tex]
[tex]\mathbf{F_{f_2} = 2 F_{f1}}[/tex]