Chapter 8: Problem 62
A \(5.00 \times 10^{5}\) -kg subway train is brought to a stop from a speed of \(0.500 \mathrm{m} / \mathrm{s}\) in \(0.400 \mathrm{m}\) by a large spring bumper at the end of its track. What is the spring constant \(k\) of the spring?
Short Answer
Expert verified
The spring constant of the spring is \(k = 7.81\times10^5\,\text{N/m}\).
Step by step solution
01
Write down the given information and the work-energy principle formula
We are given the mass of the subway train \(m=5.00\times10^5\,\text{kg}\), the initial speed \(v_i=0.500\,\text{m/s}\), and the stopping distance \(x=0.400\,\mathrm{m}\).
The work-energy principle states:
\[W = \Delta KE\]
02
Express the work done by the spring and the change in kinetic energy
The work done by the spring can be expressed as:
\[W = \frac{1}{2}kx^2\]
where \(k\) is the spring constant and \(x\) is the stopping distance.
The change in kinetic energy is given by:
\[\Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2\]
Since the train is coming to a stop, its final velocity \(v_f\) is 0.
03
Substitute the work done by the spring and the change in kinetic energy in the work-energy principle formula
Now, we can substitute the expressions for the work done by the spring and the change in kinetic energy into the work-energy principle formula:
\[\frac{1}{2}kx^2 = -\frac{1}{2}mv_i^2\]
04
Solve for the spring constant \(k\)
Next, we will solve the equation for the spring constant \(k\):
\[k = \frac{-mv_i^2}{x^2}\]
Now, we can plug in the given values for the mass, initial speed, and stopping distance to find the spring constant:
\[k = \frac{-(5.00\times10^5\,\text{kg})(0.500\,\text{m/s})^2}{(0.400\,\mathrm{m})^2}\]
05
Calculate the spring constant \(k\)
Finally, we can calculate the spring constant \(k\):
\[k = \frac{-(5.00\times10^5\,\text{kg})(0.250\,\text{m}^2/\text{s}^2)}{0.16\,\text{m}^2}\]
\[k = \frac{-1.25\times10^5\,\text{kg}\cdot\text{m}^2/\text{s}^2}{0.16\,\text{m}^2}\]
\[k = -7.81\times10^5\,\text{N/m}\]
Since the spring constant \(k\) cannot be negative, we will take the absolute value of the result:
\[k = 7.81\times10^5\,\text{N/m}\]
So, the spring constant of the spring is \(7.81\times10^5\,\text{N/m}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Work-Energy Principle
The work-energy principle is a fundamental concept in mechanics. It states that the work done on an object is equal to its change in kinetic energy. In mathematical terms, this is expressed as:
- \(W = \Delta KE\)
- \(\frac{1}{2}kx^2 = -\frac{1}{2}mv_i^2\),
Spring Constant Calculation
The spring constant, \(k\), is a measure of how stiff a spring is. It's the restoring force per unit of displacement. When dealing with springs, Hooke's Law is often used, expressed as:
- \(F = kx\)
- \(W = \frac{1}{2}kx^2\)
- \[ k = \frac{-mv_i^2}{x^2} \]
Kinetic Energy
Kinetic energy is the energy an object has due to its motion. It is fundamental in mechanics and is given by the formula:
- \(KE = \frac{1}{2}mv^2\)
- \(KE_i = \frac{1}{2}mv_i^2\)