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The velocity of the cylinder after it has dropped 0.5 in is 0.497 m/s.

Conservation of energy

The velocity of the cylinder can be determined by applying the principle of conservation of energy as shown below;

Ei = Ef

¹/₂k(xi)² + mgh = ¹/₂mv² + ¹/₂k(xt)²

where;

  • k is spring constant = 6 lb/in = 1050 N/m
  • m is mass of the system = 100 lb = 45.35 kg
  • xi is initial extension = 3 in = 0.076 m
  • xt is the total extension = 2xi + 0.5 in = 6.5 in = 0.165 m
  • h is height = 0.5 in = 0.0127 m

¹/₂k(xi)² = -mgh + ¹/₂mv² + ¹/₂k(xt)²

¹/₂(1050)(0.076)² = -(45.35)(9.8)(0.0127) + ¹/₂(45.35)v² + ¹/₂(1050)(0.165)²

3.032 = -5.64 + 22.68v² + 14.29

3.032 = 8.65 + 22.68v²

-5.62 = 22.68v²

|v| = 0.497 m/s

Thus, the velocity of the cylinder after it has dropped 0.5 in is 0.497 m/s.

The complete question is below:

The spring has a stiffness of 6 lb/in and the mass of the load is 100 lb.

Learn more about conservation of energy here: https://brainly.com/question/166559

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