Answer
a) $3.77\;\rm kN$
b) $28.1\;\rm m/s$
Work Step by Step
a)
Since the rolling friction in the pulley is negligible, the tension force is constant along the cable.
So we need to find the tension force on the cable that makes the net force exerted on the car in the $x$-direction (which is parallel to the inclined), while the car is applying the braking force, zero.
As we see in the first figure below, the net force exerted on the counterweight in the $x$-direction is zero (since the speed is constant ) and is given by
$$\sum F_{x,counter}=T-m_{counter}g\sin\theta_2=m_{counter}a_x=m_{counter}(0)$$
Thus;
$$T=m_{counter}g\sin\theta_2\tag 1$$
As we see in the second figure below, the net force exerted on the car in the $x$-direction is given by
$$\sum F_{x,car}=m_{car}g\sin\theta_1-T-F_{brakes}=m_{car}a_x=m_{car}(0)$$
Thus;
$$m_{car}g\sin\theta_1-T=F_{brakes}$$
Plugging $T$ from (1);
$$F_{brakes}=m_{car}g\sin\theta_1-m_{counter}g\sin\theta_2$$
Plugging the known;
$$F_{brakes}=\left[2000\cdot 9.8\sin30^\circ\right]-\left[1800\cdot 9.8\sin20^\circ\right]$$
$$F_{brakes}=\color{red}{\bf 3.77\times 10^3}\;\rm N$$
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b)
When the brakes fail, as we see in the last figure below, the only force that prevents it from falling is the tension force.
So to find the car's speed at the bottom point of the hill, we need to find its acceleration.
Both, the car and the weight, are having the same acceleration.
$$\sum F_{x,car}=m_{car}g\sin\theta_1-T=m_{car}a_x$$
$$m_{car}g\sin\theta_1-T=m_{car}a_x\tag 2$$
$$\sum F_{x,weight}=T-m_{weight}g\sin\theta_2 =m_{weight}a_x$$
Thus;
$$T=m_{weight}g\sin\theta_2+m_{weight}a_x$$
Plugging into (2);
$$m_{car}g\sin\theta_1-[m_{weight}g\sin\theta_2+m_{weight}a_x]=m_{car}a_x $$
$$m_{car}g\sin\theta_1- m_{weight}g\sin\theta_2-m_{weight}a_x =m_{car}a_x $$
Solving for $a_x$;
$$m_{car}g\sin\theta_1- m_{weight}g\sin\theta_2=m_{car}a_x+m_{weight}a_x $$
$$a_x=\dfrac{m_{car}g\sin\theta_1- m_{weight}g\sin\theta_2}{m_{car} +m_{weight}}$$
Plugging the known;
$$a_x=\dfrac{[2000\cdot 9.8\sin30^\circ ]- [1800\cdot 9.8 \sin20^\circ]}{2000 +1800}=\bf 0.99\;\rm m/s^2$$
Now we can fins the speed of the car by using the kinematic formula of velocity squared.
$$v_{fx}^2=\overbrace{v_{ix}^2}^{0} +2a_x\Delta x$$
Thus,
$$v_{fx}=\sqrt{2a_x\Delta x}=\sqrt{2(0.99)\Delta x} $$
Now we need to find $\Delta x$ which is, as we see in the last figure below, is given by
$$\sin\theta_1=\dfrac{200}{\Delta x}$$
where $d=\Delta x$
Thus;
$$\Delta x=\dfrac{200}{\sin\theta_1}=\dfrac{200}{\sin30^\circ }=\bf 400\;\rm m$$
Plugging into (3);
$$v_{fx}= \sqrt{2(0.99)(400)} =\color{red}{\bf 28.1}\;\rm m/s$$
