Answer:
Ff = 410.94 N
Explanation:
In order to calculate the friction force that the bullet experiences while it travels trough the bag, you take into account that the work done by the friction force in the bag, is equal to the change in the kinetic energy of the bullet. You use the following formula:
[tex]W_b=F_fd=\Delta K=\frac{1}{2}m(v^2-v_o^2)[/tex] (1)
Wb: work done by the friction force
Ff: friction force = ?
d: thick of the sand bag = 15cm = 0.15m
m: mass of the bullet = 37g = 37*10^-3 kg
v: bullet's speed after it go out from the sand bag = 32m/s
vo: bullet's speed before it enter to the sand bag = 66m/s
You solve the equation (1) or Ff, and replace the values of the other parameters:
[tex]F_f=\frac{m(v^2-v_o^2)}{2d}\\\\F_f=\frac{(37*10^{-3}kg)((32m/s)^2-(66m/s)^2)}{2(0.15m)}=-410.94N[/tex]
The minus sign means that the friction force is against the direction of the motion of the bullet.
The magnitude of the friction force is 410.94 N
