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Chapter 6: Problem 37

A 6.0-kg box moving at 3.0 \(\mathrm{m} / \mathrm{s}\) on a horizontal, frictionless surface runs into a light spring of force constant 75 \(\mathrm{N} / \mathrm{cm}\) . Use the work-energy theorem to find the maximum compression of the spring.

Short Answer

Expert verified
The maximum compression of the spring is approximately 0.085 meters.

Step by step solution

01

Convert Units for Spring Constant

First, convert the spring constant from newtons per centimeter to newtons per meter. The given spring constant is 75 N/cm, which is equivalent to 7500 N/m (since 1 m = 100 cm).
02

Calculate Initial Kinetic Energy

Find the initial kinetic energy of the box using the formula: \( KE = \frac{1}{2}mv^2 \). With mass \( m = 6.0 \) kg and velocity \( v = 3.0 \) m/s, the kinetic energy is \( KE = \frac{1}{2} \times 6.0 \times (3.0)^2 = 27 \) J.
03

Set Up Work-Energy Equation

According to the work-energy theorem, the initial kinetic energy will be equal to the work done in compressing the spring: \( KE = \frac{1}{2}kx^2 \). Here, \( k = 7500 \) N/m and \( x \) is the compression in meters.
04

Solve for Maximum Compression

Rearrange the equation \( \frac{1}{2}kx^2 = 27 \) to solve for \( x \):\[27 = \frac{1}{2} \times 7500 \times x^2\]\[x^2 = \frac{27 \times 2}{7500} = \frac{54}{7500}\]\[x^2 = 0.0072\]\[x = \sqrt{0.0072} \approx 0.085 \text{ m}\]Thus, the maximum compression of the spring is approximately 0.085 meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. Imagine being on a swing鈥攚hen you're moving, you have kinetic energy! The mathematical formula for kinetic energy is given by \[ KE = \frac{1}{2}mv^2 \]where:
  • \( m \) is the mass of the object (in kilograms),
  • \( v \) is the velocity (speed) of the object (in meters per second).
Kinetic energy is measured in joules (J). In our example, the kinetic energy of the box before it hits the spring is determined by its mass (6.0 kg) and its velocity (3.0 m/s). Plugging these numbers into the formula, we find that the kinetic energy is 27 J. This energy is crucial because it will be transformed by the work done on the spring when the box comes to a stop.
Spring Constant
The spring constant, often represented by \( k \), measures how stiff or strong a spring is. Think of it as the spring's resistance against being compressed or stretched. A larger spring constant indicates a stiffer spring.The units of the spring constant are newtons per meter (N/m). In the exercise, the given spring constant was 75 N/cm, which needed conversion to the standard unit N/m. Since 1 meter equals 100 centimeters, multiplying 75 by 100 gives us 7500 N/m. The spring constant plays an essential role in determining how much a spring will compress under a given force.When dealing with scenarios involving springs, such as the box running into our spring, it is crucial to have the spring constant in appropriate units. This way, you can accurately calculate other factors, such as the compression of the spring.
Maximum Compression of a Spring
The maximum compression of a spring occurs when all the kinetic energy of the moving object has been transferred to the spring, halting the motion. This is where the work-energy theorem comes into play.The work-energy theorem states that the initial kinetic energy of the object is transformed into potential energy in the compressed spring. The relationship can be captured by the formula:\[ KE = \frac{1}{2}kx^2 \]where:
  • \( KE \) is the initial kinetic energy,
  • \( k \) is the spring constant,
  • \( x \) is the compression distance in meters.
In this case, the initial kinetic energy of the box is 27 J, and it comes to a stop by compressing the spring. So, we set up our equation:\[ 27 = \frac{1}{2} \times 7500 \times x^2 \]Solving for \( x \) leads us to a compression of approximately 0.085 meters. This simple calculation shows how the energy from the moving box is used to compress the spring to its maximum potential.

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Most popular questions from this chapter

A luggage handler pulls a 20.0 -kg suitcase up a ramp inclined at \(25.0^{\circ}\) above the horizontal by a force \(\vec{\boldsymbol{F}}\) of magnitude 140 \(\mathrm{N}\) that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is \(\mu_{\mathrm{k}}=0.300\) . If the suitcase travels 3.80 \(\mathrm{m}\) along the ramp, calculate (a) the work done on the suitcase by the force \(\vec{\boldsymbol{F}} ;\) (b) the work done on the suitcase by the gravitational force; (c) the work done on the suitcase by the normal force; (d) the work done on the suitcase by the friction force; (e) the total work done on the suitcase. (f ) If the speed of the suitcase is zero at the bottom of the ramp, what is its speed after it has traveled 3.80 \(\mathrm{m}\) along the ramp?

BIO Should You Walk or Run? It is 5.0 \(\mathrm{km}\) from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10 \(\mathrm{km} / \mathrm{h}\) (which uses up energy at the rate of 700 \(\mathrm{W}\) ), or you could walk it leisurely at 3.0 \(\mathrm{km} / \mathrm{h}\) (which uses energy at 290 \(\mathrm{W}\) W). Which choice would burn up more energy, and how much energy (in joules) would it burn? Why is it that the more intense exercise actually burns up less energy than the less intense exercise?

Six diesel units in series can provide 13.4 \(\mathrm{MW}\) of power to the lead car of a freight train. The diesel units have total mass \(1.10 \times 10^{6} \mathrm{kg}\) . The average car in the train has mass \(8.2 \times 10^{4} \mathrm{kg}\) and requires a horizontal pull of 2.8 \(\mathrm{kN}\) to move at a constant 27 \(\mathrm{m} / \mathrm{s}\) on level tracks. (a) How many cars can be in the train under these conditions? (b) This would leave no power for accelerating or climbing hills. Show that the extra force needed to accelerate the train is about the same for a \(0.10-\mathrm{m} / \mathrm{s}^{2}\) acceleration or a 1.0\(\%\) slope (slope angle \(\alpha=\arctan 0.010 )\) . (c) With the 1.0\(\%\) slope, show that an extra 2.9 \(\mathrm{MW}\) of power is needed to maintain the \(27-\mathrm{m} / \mathrm{s}\) speed of the diesel units. (d) With 2.9 \(\mathrm{MW}\) less power available, how many cars can the six diesel units pull up a 1.0\(\%\) slope at a constant 27 \(\mathrm{m} / \mathrm{s} ?\)

You are asked to design spring bumpers for the walls of a parking garage. A freely rolling \(1200-\mathrm{kg}\) car moving at 0.65 \(\mathrm{m} / \mathrm{s}\) is to compress the spring no more than 0.090 \(\mathrm{m}\) before stopping. What should be the force constant of the spring? Assume that the spring has negligible mass.

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