Answer
The kinetic energy of the two boxes just before they strike the floor is $~~29.4~J$
Work Step by Step
We can use conservation of momentum to find the speed of the two boxes just after the collision:
$p_f = p_i$
$(3.2~kg+2.0~kg)~v_f = (3.2~kg)(3.0~m/s)$
$v_f = \frac{(3.2~kg)(3.0~m/s)}{5.2~kg}$
$v_f = 1.85~m/s$
Note that this will be the horizontal component of velocity just before the boxes strike the floor.
We can find the vertical component of the velocity just before hitting the floor:
$v_y^2 = v_{y0}^2+2ah$
$v_y^2 = 0+2ah$
$v_y = \sqrt{2ah}$
$v_y = \sqrt{(2)(9.8~m/s^2)(0.40~m)}$
$v_y = 2.8~m/s$
We can find the magnitude of the velocity just before hitting the floor:
$v = \sqrt{v_x^2+v_y^2}$
$v = \sqrt{(1.85~m/s)^2+(2.8~m/s)^2}$
$v = 3.36~m/s$
We can find the kinetic energy of the two boxes just before they strike the floor:
$K = \frac{1}{2}mv^2$
$K = \frac{1}{2}(5.2~kg)(3.36~m/s)^2$
$K = 29.4~J$
The kinetic energy of the two boxes just before they strike the floor is $~~29.4~J$
