Question

A 28 g bullet pierces a sand bag 30 cmthick. If the initial bullet velocity was 55m/s and it emerged from the sandbag with18 m/s what is the magnitude of the frictionforce (assuming it to be constant) the bullet experienced while ittraveled through the bag?
a. 130 N
b. 1.3 N
c. 13 N
d. 38 N

Answer 1

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 $m = 28 \times 10^{-3}$ kg

Thickness of sand bag $d = 30 \times 10^{-2}$ m

Initial bullet velocity $v_{i} = 55 \frac{m}{s}$

Final bullet velocity $v_{f} = 18 \frac{m}{s}$

According to the work energy theorem,

Work done by friction force is equal to change in kinetic energy.

   $\frac{1}{2} m v_{i} ^{2} - \frac{1}{2} m v_{f } ^{2} = f_{s} d$

Here we have to find friction force $f_{s}$

  $\frac{1}{2} \times 0.028 (3025-324) = f_{s} \times 0.30$

  $f_{s} = 126.046$ N

Here not any option is matching but only (A) option is near to our answer.

Therefore, the frictional force is 126.046 N