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.
[tex] \frac{1}{2}(0.015)(39)^2 = \frac{1}{2}(0.015)(27)^2 +U_{L}[/tex]
The "U" is the energy lost when the bullet went through the bag. Solving for U, we get:
[tex]U= \frac{1}{2}(0.015)(39)^2- \frac{1}{2}(0.015)(27)^2= 5.94J [/tex]
Next we know the bag is 48cm thick and that F*d=W, rearranging this equation we get:
[tex]F= \frac{W}{d} = \frac{(5.94J)}{0.48m} = 12.375N[/tex]
F = 12.375N
