Question

A 15-g bullet pierces a sand bag that is 48cm thick. The initial speed of the bullet is 39 m/s and it emerged from the sandbag at speed 27 m/s. What is the average magnitude of the friction force that slowed the bullet as it traveled through the bag?

Answer 1

We can use the conservation of energy to solve this problem. 

We know the bullet goes in at 39m/s and leaves with 27m/s and some energy lost. We can model this using the kinetic equation energy.

$\frac{1}{2}(0.015)(39)^2 = \frac{1}{2}(0.015)(27)^2 +U_{L}$

The "U" is the energy lost when the bullet went through the bag. Solving for U, we get:

$U= \frac{1}{2}(0.015)(39)^2- \frac{1}{2}(0.015)(27)^2= 5.94J$

Next we know the bag is 48cm thick and that F*d=W, rearranging this equation we get:

$F= \frac{W}{d} = \frac{(5.94J)}{0.48m} = 12.375N$

F = 12.375N