Answer:
10.2 m
Step-by-step explanation:
During the collision, momentum is conserved. Assuming the student is not moving after he hits the boxes:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(50 kg) (5.8 m/s) + 0 = 0 + (65 kg) v
v = 4.46 m/s
After the collision, energy is conserved. The kinetic energy of the boxes is converted to work done by friction.
KE = W
½ mv² = Fd
½ (65 kg) (4.46 m/s)² = (382 N) d
d = 1.69 m
The boxes move 1.69 meters each time the student hits it. After 6 times, the boxes will move a total distance of 10.2 meters.