Answer
(a) Since they only need 5.0 seconds, they will have just enough time to save the day with 0.73 seconds to spare.
(b) The kinetic energy is not conserved. The kinetic energy decreases by 4760 J.
Work Step by Step
(a) We can use conservation of energy to find the speed of the wagon at the bottom of the hill.
$K = PE$
$\frac{1}{2}m_1v_1^2 = m_1gh$
$v_1 = \sqrt{2gh}$
$v_1 = \sqrt{(2)(9.80~m/s^2)(50~m~sin(6.0^{\circ}))}$
$v_1 = 10.12~m/s$
We can use conservation of momentum to find the speed of the wagon after the two people jump on.
$m_2~v_2 = m_1~v_1$
$v_2 = \frac{m_1~v_1}{m_2}$
$v_2 = \frac{(300~kg)(10.12~m/s)}{435~kg}$
$v_2 = 6.98~m/s$
We can find the time for the wagon to travel 40 meters.
$t = \frac{x}{v_2} = \frac{40~m}{6.98~m/s}$
$t = 5.73~s$
Since they only need 5.0 seconds, they will have just enough time to save the day with 0.73 seconds to spare.
(b) $K_1 = \frac{1}{2}m_1~v_1^2$
$K_1 = \frac{1}{2}(300~kg)(10.12~m/s)^2$
$K_1 = 15360~J$
$K_2 = \frac{1}{2}m_2~v_2^2$
$K_2 = \frac{1}{2}(435~kg)(6.98~m/s)^2$
$K_2 = 10600~J$
$\Delta K = K_2 - K_1$
$\Delta K = 10600~J - 15360~J$
$\Delta K = -4760~J$
The kinetic energy is not conserved. The kinetic energy decreases by 4760 J.
