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Chapter 7: Problem 23

A 2.50 -kg mass is pushed against a horizontal spring of force constant 25.0 \(\mathrm{N} / \mathrm{cm}\) on a frictionless air table. The spring is attached to the tabletop, and the mass is not attached to the spring in any way. When the spring has been compressed enough to store 11.5 \(\mathrm{J}\) of potential energy in it, the mass is suddenly released from rest. (a) Find the greatest speed the mass reaches. When does this occur? (b) What is the greatest acceleration of the mass, and when does it occur?

Short Answer

Expert verified
(a) The greatest speed is approximately 3.03 m/s at release. (b) The greatest acceleration is 96.6 m/s虏, also at release.

Step by step solution

01

Convert Units for the Spring Constant

The spring constant is given as \( 25.0 \, \text{N/cm} \). Convert it to \( \text{N/m} \) by multiplying it by 100 (since 1 m = 100 cm). \[ k = 25.0 \, \text{N/cm} \times 100 \, \text{cm/m} = 2500 \, \text{N/m} \]
02

Calculate Maximum Compression of the Spring

The potential energy stored in the spring is given as 11.5 \( \text{J} \). Use the potential energy formula for a spring to find the compression, \( x \):\[ PE = \frac{1}{2} k x^2 \]Substitute the known values:\[ 11.5 = \frac{1}{2} \, (2500) \, x^2 \]Solve for \( x \):\[ x^2 = \frac{11.5 \times 2}{2500} \]\[ x^2 = \frac{23}{2500} \]\[ x = \sqrt{\frac{23}{2500}} \approx 0.0966 \, \text{m} \]
03

Find Greatest Speed of the Mass

All the potential energy converts to kinetic energy at maximum speed. Use:\[ KE = \frac{1}{2} mv^2 \]Equate potential energy to kinetic energy:\[ 11.5 = \frac{1}{2} \, (2.50) \, v^2 \]Solve for \( v \):\[ v^2 = \frac{11.5 \times 2}{2.50} \]\[ v^2 = 9.2 \]\[ v = \sqrt{9.2} \approx 3.03 \, \text{m/s} \]The maximum speed occurs when all potential energy is converted to kinetic energy, right after release.
04

Calculate Greatest Acceleration of the Mass

Use the formula for acceleration using Hooke's law:\[ F = ma = kx \]Solve for \( a \):\[ a = \frac{kx}{m} \]Substitute the values:\[ a = \frac{2500 \, \text{N/m} \times 0.0966 \, \text{m}}{2.50 \, \text{kg}} \]\[ a = \frac{241.5}{2.50} \approx 96.6 \, \text{m/s}^2 \]The greatest acceleration occurs at the point of maximum compression (initial release).

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Key Concepts

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

Potential Energy in Springs
Potential energy is the stored energy of an object due to its position or state. In spring mechanics, when a spring is compressed or stretched, it stores potential energy. For this, we use the formula:
  • \( PE = \frac{1}{2} k x^2 \)
Here, \( k \) is the spring constant indicating the stiffness of the spring, and \( x \) is the compression or extension relative to the spring's natural length.
When the spring in the exercise was compressed, it stored 11.5 Joules. This energy is ready to convert into motion once released. Understanding potential energy helps predict how a system stores and uses energy in physics problems.
Kinetic Energy and Its Conversion
Kinetic energy refers to the energy an object has due to its motion. When the stored potential energy in a spring is released by letting go of the compression, it transforms into kinetic energy. This follows the equation:
  • \( KE = \frac{1}{2} m v^2 \)
where \( m \) is the mass and \( v \) is the velocity of the moving object.
The energy principle in this physics problem shows that when the mass is released from the compressed spring, the potential energy becomes kinetic energy, naturally causing the object to move faster. This is how we determine the greatest speed in mechanics problems.
Hooke's Law in Spring Mechanics
Hooke's Law describes how the force exerted by a spring relates to its compression or extension. The law is expressed as:
  • \( F = kx \)
where \( F \) is the force applied by the spring, and \( x \) is the displacement from its equilibrium position.
In this exercise, Hooke's Law helps in finding the maximum force that leads to the greatest acceleration. By knowing the spring constant (\( k \)) and the extent of compression (\( x \)), Hooke's Law lets us calculate these relevant forces effectively. Understanding this is crucial in solving and predicting outcomes in physics problems involving springs.
Acceleration Explained
Acceleration is the rate at which an object changes its velocity. In the context of spring dynamics, upon release, the mass experiences acceleration due to the force applied by the spring. The acceleration can be calculated using:
  • \( a = \frac{kx}{m} \)
This formula arises from combining Newton's second law \( F = ma \) with Hooke's Law \( F = kx \).
The exercise problem shows that the greatest acceleration occurs at the point of maximum spring compression. Understanding how acceleration works helps predict how quickly an object will pick up speed right after it's released. It's a fundamental concept in physics that connects force and motion.
Understanding Physics Problems
Physics problems, like the one discussed here, help illustrate abstract concepts with practical examples. They combine definitions, laws, and calculations to provide a clearer understanding. In this case, the concepts of potential and kinetic energy, Hooke's Law, and acceleration come together.
By clearly defining each piece of the problem:
  • Unit conversions are crucial for accuracy.
  • Equating energies helps determine conditions like speed and force.
  • Interpreting formulas leads to understanding changes like acceleration.
Such problems deepen our comprehension of how forces and motions interplay in the physical world, making learning dynamic and impactful.

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Most popular questions from this chapter

A 2.00 -kg block is pushed against a spring with negligible mass and force constant \(k=400 \mathrm{N} / \mathrm{m},\) compressing it 0.220 \(\mathrm{m}\) . When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope 37.0" (Fig. \(\mathrm{P7} .42\) ). (a) What is the speed of the block as it slides along the horizontal surface after having left the spring? (b) How far does the block travel up the incline before starting to slide back down?

EALE A force parallel to the \(x\) -axis acts on a particle moving along the \(x\) -axis. This force produces potential energy \(U(x)\) given by \(U(x)=\alpha x^{4},\) where \(\alpha=1.20 \mathrm{J} / \mathrm{m}^{4} .\) What is the force (magnitude and direction) when the particle is at \(x=-0.800 \mathrm{m}\) ?

CALC A cutting tool under microprocessor control has several forces acting on it. One force is \(\vec{\boldsymbol{F}}=-\alpha x y^{2} \hat{\boldsymbol{J}}\) , a force in the negative \(y\) -direction whose magnitude depends on the position of the tool. The constant is \(\alpha=2.50 \mathrm{N} / \mathrm{m}^{3} .\) Consider the displacement of the tool from the origin to the point \(x=3.00 \mathrm{m}\) .\(y=3.00 \mathrm{m} .\) (a) Calculate the work done on the tool by \(\vec{\boldsymbol{F}}\) if this displacement is along the straight line \(y=x\) that connects these two points. (b) Calculate the work done on the tool by \(\vec{\boldsymbol{F}}\) if the tool is first moved out along the \(x\) -axis to the point \(x=3.00 \mathrm{m}, y=0\) and then moved parallel to the \(y\) -axis to the point \(x=3.00 \mathrm{m}, y=3.00 \mathrm{m}\) . (c) Compare the work done by \(\vec{\boldsymbol{F}}\) along these two paths. Is \(\vec{\boldsymbol{F}}\) conservative or nonconservative? Explain.

You are designing a delivery ramp for crates containing exercise equipment. The \(1470-\mathrm{N}\) crates will move at 1.8 \(\mathrm{m} / \mathrm{s}\) at the top of a ramp that slopes downward at \(22.0^{\circ} .\) The ramp exerts a 550 -N kinetic friction force on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 8.0 m along the ramp. Once stopped, a crate must not rebound back up the ramp. Calculate the force constant of the spring that will be needed in order to meet the design criteria.

Legal Physics. In an auto accident, a car hit a pedestrian and the driver then slammed on the brakes to stop the car. During the subsequent trial, the driver's lawyer claimed that he was obeying the posted 35 -mph speed limit, but the legal speed was too high to allow him to see and react to the pedestrian in time, You have been called in as the state's expert witness. Your investigation of the accident found that the skid marks made while the brakes were applied were 280 \(\mathrm{ft}\) long, and the tread on the tires produced a coefficient of kinetic friction of 0.30 with the road. (a) In your testimony in court, will you say that the driver was obeying the posted speed? You must be able to back up your conclusion with clear reasoning because one of the lawyers will surely cross-examine you. (b) If the driver's speeding ticket were \(\$ 10\) for each mile per hour he was driving above the posted speed limit, would he have to pay a tine? If so, how much would it be?
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