Question:

A block A of weight 500 lb is attached to a spring. The spring is initially compressed 2 ft and the block is located at a height 15 ft above the ground. As the block is released, it drops vertically until the spring stretches enough to momentarily stop the block before it rebounds upward.

• a. What is the change in kinetic energy of the block between the release and its lowest point?
• b. What is the minimum spring constant required for this motion?

Answer with Detailed Explanation:

Step 1: Understanding the Energy Principle

According to the Work-Energy Principle, when only conservative forces (like gravity and spring force) act on a system, its total mechanical energy is conserved. The energy transforms between gravitational potential energy and spring potential energy.

Step 2: Kinetic Energy Analysis

• At the release point, block A is stationary, hence its initial kinetic energy is 0 ft·lb.
• At the lowest point, the block is again momentarily at rest as the spring’s force equals the weight.

Therefore, change in kinetic energy:
ΔT = T₂ − T₁ = 0 ft·lb − 0 ft·lb = 0 ft·lb

Conclusion: There is no net change in kinetic energy.

Step 3: Determining Spring Constant using Energy Conservation

• Initial spring compression, x₁ = -2 ft
• Initial height, y₁ = 15 ft
• Clearance at lowest point, y₂ = 8 ft
• Height drop, Δh = y₁ - y₂ = 7 ft
• Final spring extension, x₂ = x₁ + Δh = -2 ft + 7 ft = 5 ft

Using the energy conservation equation:

W(y₁ - y₂) = (1/2)k(x₂² - x₁²)

Substituting values:

500(15 - 8) = (1/2)k[(5)² - (-2)²]

3500 = (1/2)k(25 - 4) = (1/2)k(21)

k = (2 × 3500) / 21 = 7000 / 21 ≈ 333.33 lb/ft

Final Answers:

• Change in Kinetic Energy: 0 ft·lb
• Minimum Spring Constant: 333.33 lb/ft