Answer
$v_2 = 25 ~m/s$
$v_3 = 11 ~m/s$
$v_4 = 19 ~m/s$
Work Step by Step
At point 1, since the cart is at rest, the energy in the system is equal to the potential energy:
$E = mgy = 32 mg ~J$
We can use conservation of energy to solve this question. At every point, the sum of the kinetic energy and potential energy will equal E.
$KE_2 + PE_2 = E$
$\frac{1}{2}mv_2^2 + mgy_2 = 32mg$
$\frac{1}{2}mv_2^2 + 0 = 32mg$
$v_2^2 = 64g$
$v_2 = \sqrt{64g} = \sqrt{(64 ~m)(9.80 ~m/s^2)}$
$v_2 = 25 ~m/s$
$KE_3 + PE_3 = E$
$\frac{1}{2}mv_3^2 + mgy_3 = 32mg$
$\frac{1}{2}mv_3^2 + 26mg = 32mg$
$\frac{1}{2}mv_3^2 = 6mg$
$v_3^2 = 12g$
$v_3 = \sqrt{12g} = \sqrt{(12 ~m)(9.80 ~m/s^2)}$
$v_3 = 11 ~m/s$
$KE_4 + PE_4 = E$
$\frac{1}{2}mv_4^2 + mgy_4 = 32mg$
$\frac{1}{2}mv_4^2 + 14mg = 32mg$
$\frac{1}{2}mv_4^2 = 18mg$
$v_4^2 = 36g$
$v_4 = \sqrt{36g} = \sqrt{(36 ~m)(9.80 ~m/s^2)}$
$v_4 = 19 ~m/s$
