What is the Force Constant k of a Spring?

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SUMMARY

The discussion centers on calculating the force constant k of a spring used to stop a subway train weighing 4.50 × 105 kg, initially traveling at 0.500 m/s over a distance of 0.900 m. The correct formula for k is derived from the relationship between kinetic energy and potential energy, specifically using k = mg/x. The initial kinetic energy (KE) of the train is calculated, and the potential energy (PE) of the spring is expressed as PE = (0.5)kx2. The correct value of k is determined to be 4900000 N/m.

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  • Knowledge of spring mechanics and Hooke's Law
  • Basic algebra for solving equations
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Homework Statement



A 4.50 ✕ 10^5 kg subway train is brought to a stop from a speed of 0.500 m/s in 0.900 m by a large spring bumper at the end of its track. What is the force constant k of the spring?

M- 4.50E5 kg
IV- .5m/s
FV- 0m/s

Homework Equations


k=mg/x
F=-kx
PE = (.5)kx^2

The Attempt at a Solution


k=(450000kg)(9.81m/s^2)/ .9 = 4900000, which isn't the right answer. Can someone help me out?
 
Last edited:
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Have you read anything about the concept of potential energy of a spring?This question is directly based on that..
The kinetic energy of the train is being converted in the potential energy of spring?
 
Well I know PE = (.5)kx^2, but I am afraid I don't know how to plug in the numbers.
 
What's the intial kinetic energy (KE) of the train?
KE will = 0 when v = 0 (why is this true?)
Due to conservation of potential energy, initial(KE+PE) = final(KE+ E)
 

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