Respuesta :
Answer:
Here not any option is matching but only (A) option is near to our answer.
The frictional force is 126.046 N
Explanation:
Given:
Mass of bullet [tex]m = 28 \times 10^{-3}[/tex] kg
Thickness of sand bag [tex]d = 30 \times 10^{-2}[/tex] m
Initial bullet velocity [tex]v_{i} = 55 \frac{m}{s}[/tex]
Final bullet velocity [tex]v_{f} = 18 \frac{m}{s}[/tex]
According to the work energy theorem,
Work done by friction force is equal to change in kinetic energy.
[tex]\frac{1}{2} m v_{i} ^{2} - \frac{1}{2} m v_{f } ^{2} = f_{s} d[/tex]
Here we have to find friction force [tex]f_{s}[/tex]
[tex]\frac{1}{2} \times 0.028 (3025-324) = f_{s} \times 0.30[/tex]
[tex]f_{s} = 126.046[/tex] N
Here not any option is matching but only (A) option is near to our answer.
Therefore, the frictional force is 126.046 N
