# A 5.8 \times 10^5 kg train is brought to a stop from a speed of 0.85 m/s in 0.57 m by a large...

## Question:

A {eq}5.8 \times 10^5 {/eq} kg train is brought to a stop from a speed of 0.85 m/s in 0.57 m by a large spring bumper at the edge of its track. What is the force constant k of the spring in N/m?

## Elastic Energy of Spring:

**SPRING**

Spring is an elastic device that, when compressed or stretched, can sore in itself elastic potential energy.

**SPRING CONSTANT OR FORCE CONSTANT**

The compression or elongation of spring is directly proportional to the external force or in other words, the restoring force of spring is directly proportional to the amount of compression or elongation but the restoring force and compression or elongation are opposite to each other in direction and the restoring force is equal to the external force in magnitude.

If an external longitudinal force {eq}F {/eq} elongated a spring by {eq}x {/eq}, then by Hooke's Law, we write:

{eq}\displaystyle{F \propto x} {/eq}.

Therefore, we put a constant and write:

{eq}\displaystyle{F = kx} {/eq}.

{eq}k {/eq} is proportionality constant called the spring constant or the force constant of the spring.

If {eq}x = 1 {/eq}, then {eq}k = F {/eq}.

Thus, the spring constant or the force constant of a spring can be defined as the force required for a unit elongation or compression of the spring.

**ELASTIC POTENTIAL ENERGY OF SPRING**

The total work done in elongating a spring of force constant {eq}k {/eq} by {eq}x {/eq} will be given by:

{eq}\displaystyle{\begin{align*} W &= \int^{x}_{0}F(x)dx\\ W &= \int^{x}_{0}kx dx\\ W &= \dfrac 12kx^2\\ \end{align*}} {/eq}

As the work is done against the restoring force of the spring which is a conservative force, the elastic potential energy of the spring will be equal to the work done against it and hence can be given by:

{eq}\displaystyle{\begin{align*} U &= W\\ U &= \dfrac 12kx^2\\ \end{align*}} {/eq}

## Answer and Explanation:

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View this answer**GIVEN**

- The mass of the train: {eq}m = 5.8\times 10^5 \ \rm kg {/eq}.

- The initial speed of the train: {eq}v = 0.85 \ \rm m/s {/eq}.

- The distance...

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Chapter 17 / Lesson 11In this lesson, you'll have the chance to practice using the spring constant formula. The lesson includes four problems of medium difficulty involving a variety of real-life applications.