Physics Problem:
Q: A block is pushed against a spring with a spring constant of 12,000 N/m and compressed by 5 cm. The block (mass = 0.476 kg) is then released, slides over a rough surface (coefficient of friction = 0.5) for a distance of 1.3 m, and then flies off horizontally, falling from a height of 2.3 m. What is the magnitude of the velocity of the block just before it hits the ground? Assume g = 9.8 m/s².
Detailed Solution:
Step 1: Energy Stored in the Spring
The elastic potential energy in a compressed spring is given by:
Espring = ½ kx²
k = 12,000 N/m, x = 0.05 m
Espring = 0.5 × 12,000 × (0.05)² = 15 J
Step 2: Work Done Against Friction
Friction does negative work and removes energy from the system:
Wf = μmgd
μ = 0.5, m = 0.476 kg, d = 1.3 m
Wf = 0.5 × 0.476 × 9.8 × 1.3 ≈ 3.027 J
Step 3: Kinetic Energy After Friction
KE = Espring – Wf
KE = 15 – 3.027 = 11.973 J
Step 4: Horizontal Velocity (vx)
KE = ½ mv² → solve for v:
vx = √(2 × 11.973 / 0.476) ≈ 7.09 m/s
Step 5: Vertical Velocity from Free Fall (vy)
vy = √(2gh)
h = 2.3 m
vy = √(2 × 9.8 × 2.3) ≈ 6.71 m/s
Step 6: Total Impact Speed
Combine horizontal and vertical components using the Pythagorean theorem:
v = √(vx² + vy²)
v = √(7.09² + 6.71²) ≈ √(50.3 + 45.0) ≈ √95.3 ≈ 9.76 m/s
Final Answer:
The velocity of the block as it strikes the ground is approximately 9.76 m/s.