Due to the total mechanical energy is conserved, you have that the potential energy of the cart at 140 m must be equal to the sum of the potential energy and kinetic energy of the cart at 95m.
Then, you can write:
[tex]\begin{gathered} K_A=K_B+U_B \\ m\cdot g\cdot h_A=\frac{1}{2}mv^2_B+m\cdot g\cdot h_B \end{gathered}[/tex]where
hA = 140m
hB = 95 m
g = 9.8 m/s^2
m = 950 kg
cancel out the mass m in the previous equation, and solve for vB:
[tex]\begin{gathered} v_B=2g\cdot h_A-2g\cdot h_B \\ v_B=\sqrt[]{2(\frac{9.8m}{s^2})(140m)-2(\frac{9.8m}{s^2})(95m)} \\ v_B=29.7\frac{m}{s} \end{gathered}[/tex]Hence, when the cart reaches the point B, its speed is approximately 29.7 m/s
