Answer:
The magnitude of the average frictional force on the block is 2 N.
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 :
[tex]v^2-u^2=2ad[/tex]
[tex]-u^2=2ad[/tex]
[tex]a=\dfrac{-u^2}{2d}\\\\a=\dfrac{-(10\ m/s)^2}{2\times 50\ m}\\\\a=-1\ m/s^2[/tex]
The frictional force on the block is given by the formula as :
F = ma
[tex]F=2\ kg\times (-1)\ m/s^2\\\\F=-2\ N[/tex]
|F| = 2 N
So, the magnitude of the average frictional force on the block is 2 N. Hence, this is the required solution.
