Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 21

How much work does it take to stretch a spring with \(k=200 \mathrm{N} / \mathrm{m}\) (a) \(10 \mathrm{cm}\) from equilibrium and (b) from \(10 \mathrm{cm}\) to \(20 \mathrm{cm}\) from equilibrium?

Short Answer

Expert verified
The work done to stretch the spring 10 cm from equilibrium is 1 J, and the additional work done to stretch the spring from 10 cm to 20 cm from equilibrium is 3 J

Step by step solution

01

Convert cm to m

To keep the units consistent, the spatial measures must be in the same units. So, convert 10 cm and 20 cm into meters. The conversion factor is 1 m = 100 cm. So, \(10 \, \text{cm} = 0.1 \, \text{m}\) and \(20 \, \text{cm} = 0.2 \, \text{m}\).
02

Calculate the work done for 10 cm stretch

Using the formula for work done in stretching or compressing a spring, which is given by \( \frac{1}{2} k x^{2}\), where k is the spring constant and x is the amount the spring is stretched or compressed, calculate the work done. Plugging in the given values gives \( \frac{1}{2} \times 200 \, \text{N/m} \times (0.1 \, \text{m})^{2} = 1 \, \text{J}\).
03

Calculate the work done to stretch the spring from 10 cm to 20 cm

To find this, you need to use the same formula and find the work done to stretch the spring to 20 cm, and then subtract from it the work done to stretch the spring to 10 cm, which is already calculated. So the work done to stretch the spring to 20 cm is \( \frac{1}{2} \times 200 \, \text{N/m} \times (0.2 \, \text{m})^{2} = 4 \, \text{J}\). The additional work done to go from 10 cm to 20 cm is \(4 \, \text{J} - 1 \, \text{J} = 3 \, \text{J}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Hooke's Law
Understanding Hooke's Law is key to solving spring-related physics problems. It is the principle that states that the force needed to extend or compress a spring by some distance is proportional to that distance. Mathematically, it is often written as:
\[ F = -kx \]
where \( F \) is the force applied, \( k \) is the spring constant indicating the stiffness of the spring, and \( x \) represents the displacement of the spring from its equilibrium position. The negative sign denotes that the force exerted by the spring is in the opposite direction to the displacement, as the spring tends to return to its original length.
Spring Force
The spring force is pivotal in many mechanical systems and follows from Hooke's Law. When a spring is either compressed or stretched, the force it exerts is equal to the spring constant \( k \) multiplied by the displacement \( x \). This restoring force is what makes springs useful in a variety of applications from clocks to car suspensions. The real-world implication of this force is that the harder the spring (higher spring constant), or the more you stretch it (greater displacement), the greater the force required.
Work-Energy Principle
The work-energy principle is a fundamental concept in physics that relates to how work is the measure of energy transfer when a force is applied over a distance. In the context of springs, the work done to stretch or compress a spring can be found using the formula:
\[ W = \frac{1}{2}kx^2 \]
where \( W \) is the work done, \( k \) is the spring constant, and \( x \) is the displacement from equilibrium. This formula comes from integrating Hooke's law over the distance and indicates that the work done is proportional to the square of the displacement.
Unit Conversion
Proper unit conversion is essential in physics to ensure accurate calculations. In spring constant work calculation, it’s common to convert from centimeters to meters since Spring constant \( k \) is typically expressed in units of Newtons per meter (N/m). Remember the conversion:
\[ 1 \text{m} = 100 \text{cm} \]
For example, to convert 10 centimeters to meters, divide by 100: \[ 10 \text{cm} = 10 / 100 = 0.1 \text{m} \] Neglecting to convert units can lead to errors in calculations, as the spring constant's unit (N/m) needs to be compatible with the displacement unit (m) for the variables in both the Hooke's law and work equations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You're trying to decide whether to buy an energy-efficient \(225-\mathrm{W}\) refrigerator for \(\$ 1150\) or a standard \(425-\mathrm{W}\) model for \(\$ 850 .\) The standard model will run \(20 \%\) of the time, but better insulation means the energy-efficient model will run \(11 \%\) of the time. If electricity costs 9.5? \(\mathrm{kW} \cdot \mathrm{h}\), how long would you have to own the energy-efficient model to make up the difference in cost? Neglect interest you might earn on your money.

An elevator of mass \(m\) rises a vertical distance \(h\) with upward acceleration equal to one-tenth \(g .\) Find an expression for the work the elevator cable does on the elevator.

The force exerted by a rubber band is given approximately by $$F=F_{0}\left[\frac{L_{0}-x}{L_{0}}-\frac{L_{0}^{2}}{\left(L_{0}+x\right)^{2}}\right]$$ where \(L_{0}\) is the unstretched length, \(x\) is the stretch, and \(F_{0}\) is a constant. Find the work needed to stretch the rubber band a distance \(x\)

A typical human diet is "2000 calories" per day, where the "calorie" describing food energy is actually 1 kilocalorie. Express 2000 kcal/day in watts.

A force \(\vec{F}\) acts in the \(x\) -direction, its magnitude given by \(F=a x^{2}\) where \(x\) is in meters and \(a=5.0 \mathrm{N} / \mathrm{m}^{2} .\) Find the work done by this force as it acts on a particle moving from \(x=0\) to \(x=6.0 \mathrm{m}\)
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Electricity and Magnetism

Read Explanation

Atoms and Radioactivity

Read Explanation

Geometrical and Physical Optics

Read Explanation

Kinematics Physics

Read Explanation

Torque and Rotational Motion

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free