91Ó°ÊÓ

The mass of the cable car $m_c=2000kg$

The mass of the counterweight$m_w=1800kg$

The force acting on the counterweight:

Sliding of the counterweight on a $20°$inclined plane

Tension acting upward

When the cable car is descending, the counterweight moves upward so the forces acting on the counterweight during the period are given by.

localid="1649649616025" $$\begin{gathered} mg\sin \left(20°\right)-T=0 \\ T=mg\sin \left(20°\right) \\ =\left(1800kg\right)\left(9.8m/s^2\right)\left(\sin \left(20°\right)\right) \\ =6033.23N \end{gathered}$$

The force acting on the cable car

Forward sliding of the cable car on a $30°$inclined plane

Tension acting upward

Breaking force, $F_b$

localid="1649649649653" $$\begin{gathered} m_{c}g\sin \left(30°\right)-T-F_b=0 \\ {F}_b=m_{c}g\sin \left(30°\right)-T \\ =\left[\left(2000kg\right)\left(9.8m/s^2\right)\left(\sin \left(30°\right)\right)\right]-\left(6033.23N\right) \\ =3766.77N \end{gathered}$$

The braking force required is $3766.77N$

The mass of the cable car$m_c=2000kg$

The mass of the counterweight $m_w=1800kg$

The net force on the counterweight

$$\begin{gathered} T-m_{c}g\sin \left(20°\right)=m_{c}a \\ T=m_{c}g\sin \left(20°\right)+m_{c}a \end{gathered}$$

The net force on the cable car is

localid="1649649718576" $$\begin{gathered} m_{c}g\sin \left(30°\right)-T=m_{c}a \\ {m}_{c}g\sin \left(30°\right)-m_{w}a+m_{w}g\sin \left(20°\right)=m_{c}a \\ g\left(m_c\sin \left(30°\right)+m_w\sin \left(20°\right)\right)=a\left(m_c+m_w\right) \\ a=\frac{g\left(m_c\sin \left(30°\right)+m_w\sin \left(20°\right)\right)}{\left(m_c+m_w\right)} \\ a=\frac{\left(9.8m/s^2\right)\left[\left(2000kg\right)×\sin \left(30°\right)+\left(1800kg\right)×\sin \left(20°\right)\right]}{\left(2000kg+1800kg\right)} \\ =0.99m/s^2 \end{gathered}$$

Now the distance traveled by car can be calculated by using trigonometry

$$\begin{gathered} \sin \left(30°\right)=\frac{200}{x} \\ x=400m \end{gathered}$$

localid="1649649896021" $$\begin{gathered} {v_f}^2={v_i}^2+2ax \\ {v}_f=\sqrt{2\left(0.99m/s^2\right)\left(400m\right)} \\ =28.28m/s \end{gathered}$$

The car's speed at the bottom of the hill is$28.28m/s$