$$W = Fd\cos\theta = (75 \text{ N})(8 \text{ m})\cos37° = 466.51 \text{ J}$$
The work done by the force of kinetic friction in opposing the motion is:
$$W_f = -f_kd = -(25 \text{ N})(8 \text{ m}) = -200 \text{ J}$$
The change in the kinetic energy of the block is:
$$\Delta K = K_f - K_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$
We can use the conservation of energy to relate the work done by the forces to the change in kinetic energy:
$$W + W_f = \Delta K$$
Substituting in the values we have calculated, we get:
$$466.51 \text{ J} - 200 \text{ J} = \frac{1}{2}(6 \text{ kg})v_f^2 - \frac{1}{2}(6 \text{ kg})(2 \text{ m/s})^2$$
Solving for $v_f$, we get:
$$v_f = 5.24 \text{ m/s}$$
Therefore, the speed of the block at the end of the 8 m displacement is 5.24 m/s.
