Question

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

Answer 1

Answer:

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

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

$F = (0.052)(-1128.7)$

$F = -58.7 N$