Problem 6
Question
(II) A \(1200-\mathrm{kg}\) car rolling on a horizontal surface has speed \(v=75 \mathrm{km} / \mathrm{h}\) when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.2 \(\mathrm{m} .\) What is the spring stiffness constant of the spring?
Step-by-Step Solution
Verified Answer
The spring stiffness constant is approximately 107,706.49 N/m.
1Step 1: Convert Speed from km/h to m/s
First, we need to convert the car's speed from kilometers per hour to meters per second. The conversion factor is \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\). Therefore, \(75 \text{ km/h} = \frac{75}{3.6} \approx 20.83 \text{ m/s}\).
2Step 2: Calculate Initial Kinetic Energy
The car's initial kinetic energy is given by \( KE = \frac{1}{2}mv^2 \), where \( m = 1200 \text{ kg} \) and \( v = 20.83 \text{ m/s} \) (the converted speed). Substituting these values gives \( KE = \frac{1}{2} \times 1200 \times (20.83)^2 \approx 260,163.89 \text{ J} \).
3Step 3: Use Energy Conservation Principle
When the car hits the spring, its kinetic energy is transformed into elastic potential energy in the spring. Using the conservation of energy, we know that \( \frac{1}{2} k x^2 = 260,163.89 \text{ J} \), where \( x = 2.2 \text{ m} \) is the compression distance and \( k \) is the unknown spring constant.
4Step 4: Solve for the Spring Constant \( k \)
Rearrange the equation from the previous step to solve for \( k \): \( k = \frac{2 \times 260,163.89}{(2.2)^2} \). Calculate this to find \( k \approx 107,706.49 \text{ N/m} \).
Key Concepts
Kinetic EnergyEnergy ConservationSpring Constant
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It's a fundamental concept in physics, especially when dealing with moving objects, like cars. The formula for kinetic energy \(KE\) is given by \(KE = \frac{1}{2}mv^2\), where \(m\) represents the mass of the object in kilograms, and \(v\) is its velocity in meters per second. This means the energy depends on both the mass and the speed squared of the object.
For instance, when a 1200 kg car travels at a speed of 20.83 m/s, its kinetic energy is calculated as \(KE = \frac{1}{2} \times 1200 \times (20.83)^2 \approx 260,163.89 \text{ J}\).
For instance, when a 1200 kg car travels at a speed of 20.83 m/s, its kinetic energy is calculated as \(KE = \frac{1}{2} \times 1200 \times (20.83)^2 \approx 260,163.89 \text{ J}\).
- This highlights how significant speed is in determining kinetic energy, as even a small increase in speed results in a large increase in energy.
Energy Conservation
Energy conservation is a powerful principle in physics that states energy cannot be created or destroyed, only transformed from one form to another. In the problem at hand, the kinetic energy of the moving car is completely converted into potential energy when it compresses the spring.
This transformation is expressed as \(\frac{1}{2}mv^2 = \frac{1}{2} k x^2\), where \(k\) is the spring constant, and \(x\) is the distance the spring is compressed, in meters.
This transformation is expressed as \(\frac{1}{2}mv^2 = \frac{1}{2} k x^2\), where \(k\) is the spring constant, and \(x\) is the distance the spring is compressed, in meters.
- The initial kinetic energy of the car, 260,163.89 J, is completely transformed into potential energy of the spring once the car comes to a stop.
- No energy is lost, illustrating the conservation principle.
Spring Constant
The spring constant, represented by the symbol \(k\), is a measure of a spring's stiffness. It indicates how much force is needed to compress or extend the spring by a certain distance. The formula used is \(F = kx\), which relates force \(F\), the spring constant \(k\), and the displacement \(x\).
In the exercise, the spring constant is found using the potential energy relation \(\frac{1}{2} k x^2\).
In the exercise, the spring constant is found using the potential energy relation \(\frac{1}{2} k x^2\).
- The energy stored in the spring when compressed by 2.2 meters is equal to the car's initial kinetic energy.
- By rearranging the formula to solve for \(k\), we achieve \(k = \frac{2 \times 260,163.89}{(2.2)^2} \).
Other exercises in this chapter
Problem 5
(II) A \(1.60-\mathrm{m}\) tall person lifts a 1.95 -kg book off the ground so it is 2.20 \(\mathrm{m}\) above the ground. What is the potential energy of the b
View solution Problem 6
(II) A 1200 -kg car rolling on a horizontal surface has speed \(v=75 \mathrm{~km} / \mathrm{h}\) when it strikes a horizontal coiled spring and is brought to re
View solution Problem 7
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(1I) A particular spring obeys the force law \(\vec{\mathbf{F}}=\) \(\left(-k x+a x^{3}+b x^{4}\right) \hat{\mathbf{i}}\) (a) Is this force conservative? Explai
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