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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 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
The greatest speed is 3.03 m/s when the spring is at equilibrium. The greatest acceleration is 96 m/s虏 when the spring is maximally compressed.

Step by step solution

01

Understand the Potential Energy Stored in the Spring

The potential energy stored in a compressed or stretched spring is given by \[ U = \frac{1}{2} k x^2 \]where \( U \) is the potential energy, \( k \) is the spring constant, and \( x \) is the compression of the spring. We're given that \( U = 11.5 \, \text{J} \), and \( k = 25.0 \, \text{N/cm} \) or \( 2500 \, \text{N/m} \) after converting to SI units.
02

Solve for the Compression of the Spring

Rearrange the equation for potential energy to solve for \( x \):\[ 11.5 = \frac{1}{2} (2500) x^2 \]\[ x^2 = \frac{11.5 \times 2}{2500} = 0.0092 \]\[ x = \sqrt{0.0092} \approx 0.096 \, \text{m} \].The spring is compressed 0.096 meters.
03

Calculate the Maximum Speed of the Mass

The work done by the spring or the potential energy stored in the spring converts entirely into kinetic energy when the spring is fully relaxed. The kinetic energy (KE) is \[ KE = \frac{1}{2} mv^2 \]where \( m = 2.50 \, \text{kg} \) is the mass. Set potential energy equal to kinetic energy:\[ 11.5 = \frac{1}{2} \times 2.50 \times v^2 \]Solving for \( v \) gives:\[ 11.5 = 1.25v^2 \]\[ v^2 = \frac{11.5}{1.25} = 9.2 \]\[ v = \sqrt{9.2} \approx 3.03 \, \text{m/s} \].The greatest speed is 3.03 m/s and this occurs when the spring returns to its equilibrium position.
04

Calculate the Maximum Acceleration of the Mass

The maximum acceleration happens when the force is maximum, which occurs when the spring is compressed the most (at the start). Force \( F \) exerted by the spring is given by Hooke's Law: \[ F = kx \]The acceleration \( a \) is given by \[ a = \frac{F}{m} \]Substitute maximally compressed \( x = 0.096 \, \text{m} \):\[ F = 2500 \times 0.096 = 240 \, \text{N} \]\[ a = \frac{240}{2.50} = 96 \, \text{m/s}^2 \].The greatest acceleration is 96 m/s虏, and it occurs when the spring is maximally compressed.

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Key Concepts

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

Potential Energy
Potential energy is the stored energy in an object due to its position or configuration. In the context of springs, it refers to the energy stored when a spring is compressed or stretched. The formula for potential energy in a spring is given by \[ U = \frac{1}{2} k x^2 \] where
  • \( U \) is the potential energy, measured in joules (J),
  • \( k \) is the spring constant, representing the stiffness of the spring, measured in newtons per meter (N/m),
  • \( x \) is the displacement from the spring's equilibrium position, measured in meters (m).
The more a spring is compressed or stretched, the more potential energy it stores. In the given problem, the spring has been compressed enough to store 11.5 J of potential energy. This energy is later converted to kinetic energy when the spring returns to its equilibrium position.
Hooke's Law
Hooke's Law describes the behavior of springs. It states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. Mathematically, it is expressed as: \[ F = kx \] where
  • \( F \) is the force applied by the spring, measured in newtons (N),
  • \( k \) is the spring constant, an indicator of how stiff the spring is,
  • \( x \) is the displacement from equilibrium, measured in meters (m).
In the problem, when the spring is initially compressed by 0.096 meters, it exerts a force, calculated using Hooke's Law, that pushes the mass. This force is maximum when the spring is maximally compressed.
Kinetic Energy
Kinetic energy is the energy of motion. When an object is moving, it possesses kinetic energy, which is given by the formula: \[ KE = \frac{1}{2} mv^2 \] where
  • \( KE \) is the kinetic energy, measured in joules (J),
  • \( m \) is the mass of the object, measured in kilograms (kg),
  • \( v \) is the velocity of the object, measured in meters per second (m/s).
In this exercise, the potential energy stored in the spring is converted into kinetic energy, propelling the mass. This happens as the spring releases its stored energy, resulting in the greatest speed (3.03 m/s) of the mass when the spring returns to its equilibrium position.
Spring Constant
The spring constant, denoted as \( k \), is a key property of a spring that defines its stiffness. It is a measure of the force required to compress or stretch the spring by a unit length. The unit for the spring constant is newtons per meter (N/m). In our exercise, the spring constant is given as 25.0 N/cm, which is converted to 2500 N/m for calculation purposes. This conversion is necessary because working with SI units (meters) simplifies mathematical calculations and ensures uniformity in physics equations. The spring constant plays a crucial role in determining the behavior of the spring according to Hooke's Law, as well as in calculating the potential energy stored when the spring is compressed.

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Most popular questions from this chapter

A \(0.150-\mathrm{kg}\) block of ice is placed against a horizontal, compressed spring mounted on a horizontal tabletop that is 1.20 \(\mathrm{m}\) above the floor. The spring has force constant 1900 \(\mathrm{N} / \mathrm{m}\) and is initially compressed 0.045 \(\mathrm{m}\) . The mass of the spring is negligible. The spring is released, and the block slides along the table, goes off the edge, and travels to the floor. If there is negligible friction between the block of ice and the tabletop, what is the speed of the block of ice when it reaches the floor?

A certain spring is found not to obey Hooke's law; it exerts a restoring force \(F_{x}(x)=-\alpha x-\beta x^{2}\) if it is stretched or compressed, where \(\alpha=60.0 \mathrm{N} / \mathrm{m}\) and \(\beta=18.0 \mathrm{N} / \mathrm{m}^{2} .\) The mass of the spring is negligible. (a) Calculate the potentinl-energy function \(U(x)\) for this spring. Let \(U=0\) when \(x=0\) (b) An object with mass 0.900 \(\mathrm{kg}\) on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 \(\mathrm{m}\) to the right (the \(+x\) -direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 \(\mathrm{m}\) to the right of the \(x=0\) equilibrium position?

When it is burned, 1 gallon of gasoline produces \(1.3 \times 10^{8} \mathrm{J}\) of energy. A \(1500-\mathrm{kg}\) car accelerates from rest to 37 \(\mathrm{m} / \mathrm{s}\) in 10 \(\mathrm{s}\) . The engine of this car is only 15\(\%\) efficient (which is typical), meaning that only 15\(\%\) of the energy from the combustion of the gasoline is used to accelerate the car. The rest goes into things like the internal kinetic energy of the engine parts as well as heating of the exhaust air and engine. (a) How many gallons of gasoline does this car use during the acceleration? (b) How many such accelerations will it take to burn up 1 gallon of gas?

The potential energy of a pair of hydrogen atoms separated by a large distance \(x\) is given by \(U(x)=-C_{6} / x^{6},\) where \(C_{6}\) is a positive constant. What is the force that one atom exerts on the other? Is this force attractive or repulsive?

A wooden block with mass 1.50 \(\mathrm{kg}\) is placed against a compressed spring at the bottom of an incline of slope \(30.0^{\circ}\) (point \(A\) ). When the spring is released, it projects the block up the incline. At point \(B,\) a distance of 6.00 \(\mathrm{mup}\) the incline from \(A\) , the block is moving up the incline at 7.00 \(\mathrm{m} / \mathrm{s}\) and is no longer in contact with the spring. The coefficient of kinetic friction between the block and the incline is \(\mu_{\mathrm{k}}=0.50 .\) The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.
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