Let's break down the problem provided in the image step by step. The setup includes:

• A 65.0 kg bucket of concrete that hangs from a friction-free pulley.
• The pulley is connected to an 80.0 kg box on a horizontal roof.
• A 50.0 kg bag of gravel rests on top of the box.
• The coefficient of static friction ($\mu_s = 0.700$) and kinetic friction ($\mu_k = 0.400$) between the box and the roof are given.

Part (a): Finding the friction force

When analyzing the friction force, we need to consider the static friction between the box with the gravel and the roof.

Forces involved:

1. Gravitational force on the gravel-bag-box system:

   • Total mass = $80.0 \, \text{kg (box)} + 50.0 \, \text{kg (gravel)} = 130.0 \, \text{kg}$
   • Normal force ($F_N$) due to gravity:
     $F_N = (130.0 \, \text{kg}) \times (9.81 \, \text{m/s}^2) = 1275.3 \, \text{N}$

2. Friction force ($F_{\text{friction}}$):

   • Using static friction:
     $F_{\text{friction}} = \mu_s \times F_N = 0.700 \times 1275.3 \, \text{N} = 892.71 \, \text{N}$

Thus, the friction force between the box and the roof is 892.71 N.

Part (b): Speed of the bucket after 2.00 m descent

Once the gravel is removed, the only mass on the box is the 80.0 kg. We'll use energy conservation to find the speed of the bucket.

Energy considerations:

• Initial gravitational potential energy of the bucket is converted into:
  • Kinetic energy of the bucket.
  • Work done against the frictional force (now with just the box on the roof).

1. Initial potential energy of the bucket: $PE_{\text{initial}} = m_{\text{bucket}} \times g \times h = (65.0 \, \text{kg}) \times (9.81 \, \text{m/s}^2) \times (2.00 \, \text{m}) = 1274.7 \, \text{J}$

2. Frictional force (with only the box now):

   • Normal force on the box alone:
     $F_N = (80.0 \, \text{kg}) \times (9.81 \, \text{m/s}^2) = 784.8 \, \text{N}$
   • Kinetic friction force:
     $F_{\text{friction}} = \mu_k \times F_N = 0.400 \times 784.8 \, \text{N} = 313.92 \, \text{N}$
   • Work done against friction over 2.00 m:
     $W_{\text{friction}} = F_{\text{friction}} \times d = 313.92 \, \text{N} \times 2.00 \, \text{m} = 627.84 \, \text{J}$

3. Final kinetic energy of the system:

   • Using energy conservation:
     $KE = PE_{\text{initial}} - W_{\text{friction}} = 1274.7 \, \text{J} - 627.84 \, \text{J} = 646.86 \, \text{J}$
   • Total mass of the system: $65.0 \, \text{kg (bucket)} + 80.0 \, \text{kg (box)} = 145.0 \, \text{kg}$
   • The speed $v$ is found from kinetic energy:
     $KE = \frac{1}{2} m_{\text{total}} v^2 \quad \Rightarrow \quad v = \sqrt{\frac{2 \times KE}{m_{\text{total}}}} = \sqrt{\frac{2 \times 646.86 \, \text{J}}{145.0 \, \text{kg}}} = 2.99 \, \text{m/s}$

Thus, the speed of the bucket after descending 2.00 m is 2.99 m/s.

Let me know if you'd like further details on any step, or if you'd like me to explore other approaches. Here are some related questions to expand your understanding:

1. What would happen to the system if the pulley had friction?
2. How would the friction force change if the box was heavier?
3. Can you derive the equation for energy conservation in a system involving friction?
4. What role does the mass of the pulley play in this scenario?
5. How would the speed of the bucket change if the descent was 3.00 m instead?

Tip: When dealing with pulley systems and friction, always break down the forces acting on each mass and use energy principles for simplification!