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

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

The frictional force is 126.046 N

Step-by-step explanation:

Given:

Mass of bullet
`m = 28 * 10^(-3)` kg

Thickness of sand bag
`d = 30 * 10^(-2)` m

Initial bullet velocity
`v_(i) = 55 (m)/(s)`

Final bullet velocity
`v_(f) = 18 (m)/(s)`

According to the work energy theorem,

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

`(1)/(2) m v_(i) ^(2) - (1)/(2) m v_(f ) ^(2) = f_(s) d`

Here we have to find friction force
`f_(s)`

`(1)/(2) * 0.028 (3025-324) = f_(s) * 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