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Example No. 10
In a worst-case design scenario, a 2000 kg elevator with broken cables is
falling at 8.0 m/s when it first contact a cushioning spring at the bottom
of the shaft. The spring is supposed to stop the elevator, compressing 3.0
m as it does so. Determine what the force constant of the spring should
be.

Respuesta :

onyebuchinnaji

The force constant of the spring is determined as 14,222.2 N/m.

Force constant of the spring

Apply the principle of conservation of energy,

K.E = U

where;

  • K.E kinetic energy of the elevator
  • U is elastic potential energy of the spring

¹/₂mv² = ¹/₂kx²

mv² = kx²

k = mv²/x²

Where;

  • m is mass of the elevator
  • v is speed
  • x is compression of the spring

k = (2000 x 8²)/(3²)

k = 14,222.2 N/m

Thus, the force constant of the spring is determined as 14,222.2 N/m.

Learn more about force constant here: https://brainly.com/question/1968517

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