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A subway train with a mass of 600,000 kg is moving at a speed of 0.5 m/s when it comes to a stop due to a large spring bumper at the end of its track. The spring bumper compresses by 0.4 meters as a result of this collision, and the energy from the train is transferred and stored in the spring. What is the spring constant of the bumper?

User DarenW
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1 Answer

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To find the spring constant of the bumper, we can equate the initial kinetic energy of the train with the potential energy stored in the compressed spring. By using the formulas for kinetic energy and potential energy in a spring, we can solve for the spring constant. The spring constant of the bumper is approximately 1,875,000 N/m.

Step-by-step explanation:

To solve this problem, we can use the principle of conservation of mechanical energy. The initial kinetic energy of the train is equal to the potential energy stored in the compressed spring bumper. The formula for potential energy in a spring is given by:

PE = 0.5kx^2

Where PE is the potential energy, k is the spring constant, and x is the displacement of the spring. In this case, the displacement is given as 0.4 m.

The initial kinetic energy of the train can be calculated using the formula:

KE = 0.5mv^2

Where KE is the kinetic energy, m is the mass of the train, and v is the velocity of the train. Plugging in the given values:

KE = 0.5 × 600,000 kg × (0.5 m/s)^2

Now, we can equate the initial kinetic energy with the potential energy:

0.5 × 600,000 kg × (0.5 m/s)^2 = 0.5k × (0.4 m)^2

Simplifying the equation:

150,000 J = 0.08k

Solving for k:

k = 150,000 J / 0.08

k ≈ 1,875,000 N/m

User Richard Lusch
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