given that initial speed of the car is
[tex]v_i = 8.9 m/s[/tex]
now after travelling the distance d = 1.8 * 10^1 m the car will stop
so here we can use kinematics to find the acceleration of car
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]0 - 8.9^2 = 2 a d[/tex]
here we have
[tex]- 79.21 = 2*(18)*a[/tex]
[tex]a = -2.2 m/s^2[/tex]
net force applied due to brakes of car is given by Newton's II law
[tex]F = ma[/tex]
here we have
mass = 1.2 * 10^3 kg
[tex]F_{net} = 1.2 * 10^3 * 2.2[/tex]
[tex]F_{net} = 2.64 * 10^3 N[/tex]
now we can say
[tex]F_{net} = F_1 + F_2[/tex]
[tex]2.64 * 10^3 = 1.8 * 10^3 + F_2[/tex]
[tex]F_2 = 8.4 * 10^2 N[/tex]
So the force applied due to brakes is given as above