According to the work-energy theorem, the net work done on an object equals its change in kinetic energy, $W_{net} = \Delta K$. The work done by friction is negative, $W_f = -1000$ J.

$$W_f = K_f - K_i$$ $$W_f = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$

Solving for the final velocity $v_f$:

$$v_f = \sqrt{v_i^2 + \frac{2W_f}{m}}$$

Substituting the values:

$$v_f = \sqrt{(6 \text{ m/s})^2 + \frac{2(-1000 \text{ J})}{100 \text{ kg}}} = \sqrt{36 - 20} \text{ m/s} = \sqrt{16} \text{ m/s}$$