A 37 g bullet pierces a sand bag 15cm thick. If the initial bullet velocity was 66 m/s and it emerged from the sandbag with 32 m/s, what is the magnitude of the friction force (assuming it to be constant) the bullet experienced while it traveled through the bag

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