Answer
The change in the mechanical energy is $~~-3750~J$
Work Step by Step
We can find the bullet's initial kinetic energy:
$K = \frac{1}{2}mv^2$
$K = \frac{1}{2}(0.030~kg)(500~m/s)^2$
$K = 3750~J$
Since the bullet comes to a stop, the change in the mechanical energy is $~~-3750~J$
