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Q22 PE

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College Physics (Urone)
Found in: Page 261

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Short Answer

A 5.00×105-kg subway train is brought to a stop from a speed of 0.500 m/s in 0.400 m by a large spring bumper at the end of its track. What is the force constant k of the spring?

The force constant of the spring is 781250 N/m.

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Step by Step Solution

Force constant

Force constant: When a spring is stretched by the application of some force, then according to the Hooke’s law the increase in the length of spring from its equilibrium length is directly proportional to the force applied.

Mathematically,

FxF=kx

Here, k is the proportionality constant known as force constant or spring constant.

Calculation of the force constant

When a spring is stretched, the change in potential energy of the spring is,

ΔPE=12kx2

Here, k is the force constant, and x is the displacement x=0.400 m.

According to the law of conservation of energy,

ΔPE=ΔKE12kx2=12mv2

Here, m is the mass of the subway train m=5.00×105 kg, and v is the initial velocity of the subway train v=0.500 m/s.

The expression for the force constant is,

k=mv2x2

Putting all known values,

k=5.00×105 kg×0.500 m/s20.400 m2=781250 N/m

Therefore, the required force constant of the spring is 781250 N/m.

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