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Chapter 6: Problem 9

A spring has a spring constant of \(80 \mathrm{~N} / \mathrm{m}\). How much potential energy does it store when stretched by \(1.0 \mathrm{~cm} ?\) a) \(4.0 \cdot 10^{-3}\) J b) \(0.40 \mathrm{~J}\) c) 80 d) \(800 \mathrm{~J}\) e) \(0.8 \mathrm{~J}\)

Short Answer

Expert verified
Answer: (a) \(4.0 \cdot 10^{-3}\) J

Step by step solution

01

Identify the given information

We are given: - spring constant (k) = \(80 \mathrm{~N/m}\) - stretch length (x) = \(1.0 \mathrm{~cm}\)
02

Convert units if necessary

Since the spring constant is given in \(\mathrm{N/m}\), we need to convert the stretch length to meters: \(x = 1.0 \mathrm{~cm} \times \dfrac{1 \mathrm{~m}}{100 \mathrm{~cm}} = 0.01 \mathrm{~m}\)
03

Apply Hooke's Law formula

Using the formula for potential energy (PE) stored in a stretched spring: \(PE = \dfrac{1}{2}kx^2\) Plug in the values of k and x: \(PE = \dfrac{1}{2}(80 \mathrm{~N/m})(0.01 \mathrm{~m})^2\)
04

Solve for the potential energy

Calculate the potential energy stored: \(PE = \dfrac{1}{2}(80 \mathrm{~N/m})(0.0001 \mathrm{m}^2) = 0.004 \mathrm{J}\)
05

Compare the result with the given options

The calculated potential energy is \(0.004 \mathrm{J}\) which is the same as \(4.0 \cdot 10^{-3} \mathrm{J}\). Therefore, the correct answer is (a) \(4.0 \cdot 10^{-3}\) J.

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Key Concepts

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

Understanding Potential Energy in Springs
When we talk about potential energy, specifically in the context of a spring, we refer to the stored energy due to its deformation. If a spring is compressed or stretched, it possesses potential energy. This is calculated using Hooke's Law for springs, with the formula: \(PE = \dfrac{1}{2}kx^2\), where
  • \(PE\) is the potential energy,
  • \(k\) is the spring constant, and
  • \(x\) is the displacement from its natural length.
This energy is what allows springs to "spring back" to their original position after being deformed, contributing to various applications like storing energy in a mechanical clock or absorbing shock in vehicle suspensions.
The Role of the Spring Constant
The spring constant, symbolized by \(k\), is a fundamental parameter in understanding a spring's behavior. It measures a spring鈥檚 stiffness 鈥� essentially, how resistant it is to being deformed.
A larger spring constant means the spring is stiffer, requiring more force to achieve the same displacement as a less stiff spring. The unit for the spring constant is Newton per meter (N/m), signifying how many newtons of force are required to stretch or compress the spring by one meter.
When solving for potential energy, knowing the spring constant is crucial, as it directly influences the magnitude of energy stored when the spring is stretched or compressed.
Why Unit Conversion Matters
Unit conversion is pivotal in physics because it ensures that all calculations are consistent and correct. In exercises involving springs, it is common for different units to be used, especially for displacement.
For example, in a spring problem with a spring constant in \(N/m\), displacement should be converted to meters to maintain consistency.
  • Converting units ensures that the calculations are valid and the formula can be applied accurately.
  • A common conversion factor is converting centimeters to meters, where 1 cm equals 0.01 m.
Therefore, before applying Hooke鈥檚 Law, always double-check that your units match those of the spring constant, as seen in this exercise where 1.0 cm was converted to 0.01 m.

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Most popular questions from this chapter

A spring with a spring constant of \(500 . \mathrm{N} / \mathrm{m}\) is used to propel a 0.500 -kg mass up an inclined plane. The spring is compressed \(30.0 \mathrm{~cm}\) from its equilibrium position and launches the mass from rest across a horizontal surface and onto the plane. The plane has a length of \(4.00 \mathrm{~m}\) and is inclined at \(30.0^{\circ} .\) Both the plane and the horizontal surface have a coefficient of kinetic friction with the mass of \(0.350 .\) When the spring is compressed, the mass is \(1.50 \mathrm{~m}\) from the bottom of the plane. a) What is the speed of the mass as it reaches the bottom of the plane? b) What is the speed of the mass as it reaches the top of the plane? c) What is the total work done by friction from the beginning to the end of the mass's motion?

You use your hand to stretch a spring to a displacement \(x\) from its equilibrium position and then slowly bring it back to that position. Which is true? a) The spring's \(\Delta U\) is positive. b) The spring's \(\Delta U\) is negative. c) The hand's \(\Delta U\) is positive. d) The hand's \(\Delta U\) is negative. e) None of the above statements is true.

A girl of mass \(49.0 \mathrm{~kg}\) is on a swing, which has a mass of \(1.0 \mathrm{~kg} .\) Suppose you pull her back until her center of mass is \(2.0 \mathrm{~m}\) above the ground. Then you let her \(\mathrm{go},\) and she swings out and returns to the same point. Are all forces acting on the girl and swing conservative?

A father exerts a \(2.40 \cdot 10^{2} \mathrm{~N}\) force to pull a sled with his daughter on it (combined mass of \(85.0 \mathrm{~kg}\) ) across a horizontal surface. The rope with which he pulls the sled makes an angle of \(20.0^{\circ}\) with the horizontal. The coefficient of kinetic friction is \(0.200,\) and the sled moves a distance of \(8.00 \mathrm{~m}\). Find a) the work done by the father, b) the work done by the friction force, and c) the total work done by all the forces.

Which of the following is not a unit of energy? a) newton-meter b) joule c) kilowatt-hour d) \(\operatorname{kg} \mathrm{m}^{2} / \mathrm{s}^{2}\) e) all of the above
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