Answer:
Frictional force will be equal to 126.04 N
Explanation:
We have given mass of the bullet m = 28 gram = 0.028 kg
Initial velocity u = 55 m /sec
Width of sand = 30 cm = 0.3 m
So initial kinetic energy [tex]E=\frac{1}{2}mu^2=\frac{1}{2}\times 0.028\times 55^2=42.35j[/tex]
Final velocity of the bullet v = 18 m /sec
So final kinetic energy [tex]E=\frac{1}{2}mv^2=\frac{1}{2}\times 0.028\times 18^2=4.536j[/tex]
So change in kinetic energy = 42.35 - 4.536 = 37.814 j
From work energy theorem this change in kinetic energy will be equal to work one by frictional force
So [tex]f\times 0.3=37.814[/tex]
[tex]f=126.04 N[/tex]