Answer:
0.65 m/min
Step-by-step explanation:
The volume of material to be removed is
80*8*10 = 650 cm^3
The tool has a diameter of 4 mm and a maximum axial cutting capacity of 50 mm, so its cross section normal to advance is
0.4*5 = 2 cm^2
If the groove have to be made in T = 5 minutes the advance speed would be
V/(S * T)
650/(2 * 5) = 65 cm/min = 0.65 m/min