Answer:
The magnitude of the average frictional force on the block is 2 N.
Step-by-step explanation:
Given that.
Mass of the block, m = 2 kg
Initial velocity of the block, u = 10 m/s
Distance, d = 50 m
Finally, it stops, v = 0
Let a is the acceleration of the block. It can be calculated using third equation of motion. It can be given by :



The frictional force on the block is given by the formula as :
F = ma

|F| = 2 N
So, the magnitude of the average frictional force on the block is 2 N. Hence, this is the required solution.