To determine how the kinetic energy of the block will change, let's analyze the forces acting on it and determine if there's any net work done in the direction of motion.

Given Data and Analysis:

• Force $F$ = 20 N, acting downward at a $60.0^\circ$ angle relative to the horizontal.
• Distance moved $d = 5.00 \, \text{m}$.
• Force $f = 3 \, \text{N}$, acting to the left (opposing the direction of motion if the block moves to the right).
• Friction Force $f_{\text{friction}} = 3.00 \, \text{N}$.
• Mass of the block $m = 5 \, \text{kg}$.

Step-by-Step Solution:

1. Resolve Force $F$ into Components: Since $F$ acts at an angle of $60^\circ$ from the horizontal:

   • Horizontal component of $F$: $F_x = F \cos(60^\circ) = 20 \times \frac{1}{2} = 10 \, \text{N}$
   • Vertical component of $F$: $F_y = F \sin(60^\circ) = 20 \times \frac{\sqrt{3}}{2} \approx 17.32 \, \text{N}$

2. Analyze the Forces in the Horizontal Direction:

   • The horizontal component $F_x = 10 \, \text{N}$ pushes the block to the right.
   • The friction force $f_{\text{friction}} = 3.00 \, \text{N}$ opposes the motion to the right.
   • Net horizontal force: $F_{\text{net}} = F_x - f_{\text{friction}} = 10 \, \text{N} - 3 \, \text{N} = 7 \, \text{N}$

3. Work Done by the Net Force:

   • Since there is a net force $F_{\text{net}} = 7 \, \text{N}$ in the horizontal direction and the block moves a distance $d = 5.00 \, \text{m}$ in this direction, work is done by this net force.
   • Work done $W = F_{\text{net}} \times d$: $W = 7 \, \text{N} \times 5.00 \, \text{m} = 35 \, \text{J}$

4. Effect on Kinetic Energy:

   • The positive work done on the block means that energy is added to the block’s motion.
   • Since work is directly related to the change in kinetic energy (by the work-energy theorem), the kinetic energy of the block will increase.

Conclusion:

The correct answer is: (A) will increase.

Would you like further details on this solution, or do you have any specific questions?

---

Here are some related questions to consider:

1. How would the answer change if the friction force was greater than the horizontal component of $F$?
2. What would happen if the force $F$ were applied at an angle greater than $60^\circ$ above the horizontal?
3. How would the kinetic energy change if the distance moved was less than 5.00 m?
4. What if there was an upward force applied instead of the downward force at $60^\circ$?
5. How does the weight of the block influence the frictional force here?

Tip: Remember, in such problems, the work done by a force component in the direction of motion directly influences the change in kinetic energy.