Answer
The total kinetic energy of the system increases by $~~40.0~J$
Work Step by Step
We can use conservation of momentum to find the velocity $v_f$ of the 5.0-kg block immediately after the collision:
$p_f = p_i$
$(5.0~kg)~v_f+(10.0~kg)(4.0~m/s) = (5.0~kg)(3.0~m/s)+(10.0~kg)(2.0~m/s)$
$(5.0~kg)~v_f = (5.0~kg)(3.0~m/s)+(10.0~kg)(2.0~m/s)-(10.0~kg)(4.0~m/s)$
$v_f = \frac{(5.0~kg)(3.0~m/s)+(10.0~kg)(2.0~m/s)-(10.0~kg)(4.0~m/s)}{5.0~kg}$
$v_f = -1.0~m/s$
We can find the total kinetic energy of the system before the collision:
$K_i = \frac{1}{2}(5.0~kg)(3.0~m/s)^2+\frac{1}{2}(10.0~kg)(2.0~m/s)^2$
$K_i = 22.5~J+20.0~J$
$K_i = 42.5~J$
We can find the total kinetic energy of the system after the collision:
$K_f = \frac{1}{2}(5.0~kg)(-1.0~m/s)^2+\frac{1}{2}(10.0~kg)(4.0~m/s)^2$
$K_f = 2.5~J+80.0~J$
$K_f = 82.5~J$
We can find the change in the kinetic energy of the system:
$\Delta K = K_f - K_i = 82.5~J - 42.5~J = 40.0~J$
The total kinetic energy of the system increases by $~~40.0~J$
