Question

In a ballistics test, a 52g bullet hits a sand bag and stops after moving 1.34 m. If the initial bullat

velocity was 55 m/s what was the magnitude of the average force of friction. ​

Answer 1

Friction force on the bullet is 58.7 N opposite to its velocity

Step-by-step explanation:

As we know that initial speed of the bullet is 55 m/s

after travelling into the sand bag by distance d = 1.34 m it comes to rest

so final speed

`v_f = 0`

now we can use kinematics top find the acceleration of the bullet

`v_f^2 - v_i^2 = 2 a d`

so we have

`0 - 55^2 = 2(a)(1.34)`

`a = -1128.7 m/s^2`

now by Newton's II law we know that

`F = ma`

so we have

`F = (0.052)(-1128.7)`

`F = -58.7 N`