Skip to content Skip to navigation

Connexions

You are here: Home » Content » Worked Example Template

Navigation

Recently Viewed

This feature requires Javascript to be enabled.
 

Worked Example Template

Module by: Katherine Fletcher. E-mail the author

Based on: Work Energy and Power - Grade 12 by Rory Adams, Free High School Science Texts Project, Mark Horner

Summary: This module is to be used as a template for worked examples. It uses examples from the Physics module it is based on, and adds some information on the efficacy of worked examples in teaching and learning.

Worked Example Background Information

What is a worked example?

Worked examples are problems that are solved using a step-by-step approach, with teaching or explanatory information for each step.

Research on Worked Examples

In 1985, Sweller and Cooper experimented with presenting worked examples to students learning algebra. They presented two groups of students with a single worked example. Then in one group, they presented several more similar worked examples, and in the second group they asked those students to solve those additional problems. Finally, they tested both groups. They found that the students who saw the extra worked examples, rather than trying to solve the extra problems, did better on the final test.

If possible, combining pictures with words is helpful. So is audio with text, video with audio, etc. The multiple channels enhance learning. Subsequent research discovered that worked examples are best for novice learners of a topic. Once the topic has been absorbed, solving actual problems may be more effective. Regardless, worked examples are an age old technique for teaching, found in most textbooks, and most classrooms.

Model Worked Examples from High School Physics

These worked examples were taken from another module published in Connexions by the Free High School Science Textbooks project, "Work, Energy, and Power - Grade 12" [link]. At the top of this module, you will see the original module listed under "This module based on ...". The FHSST books have lots of worked examples.

The format of the examples has been reworked to use the "stepwise" list type in CNXML. Feel free to borrow from the XML in these examples to create new worked examples in other modules. To see the source for the examples, you would append /source to the URL you are using to view this module, and then search for the work "exercise".

Exercise 1: Calculating Work Done I

If you push a box 20 m forward by applying a force of 15 N in the forward direction, what is the work you have done on the box?

Solution

  1. Step 1. Analyze the question to determine what information is provided:
    • The force applied is F=15 N.
    • The distance moved is s=20 m.
    • The applied force and distance moved are in the same direction. Therefore, F∥=15 N.

    These quantities are all in the correct units, so no unit conversions are required.

  2. Step 2. Analyze the question to determine what is being asked:
    • We are asked to find the work done on the box. We know from the definition that work done is W=F∥sW=F∥s
  3. Step 3. Next we substitute the values and calculate the work done:
    W = F ∥ s = ( 15 N ) ( 20 m ) = 300 J W = F ∥ s = ( 15 N ) ( 20 m ) = 300 J
    (1)

Remember that the answer must be positive as the applied force and the motion are in the same direction (forwards). In this case, you (the pusher) lose energy, while the box gains energy.

Exercise 2: Calculating Work Done II

What is the work done by you on a car, if you try to push the car up a hill by applying a force of 40 N directed up the slope, but it slides downhill 30 cm?

Solution

Steps for solving this problem
  1. Step 1. Analyze the question to determine what information is provided:
    • The force applied is F=40 N
    • The distance moved is s=30 cm. This is expressed in the wrong units so we must convert to the proper S.I. units (meters):
      100 cm = 1 m 1 cm = 1 100 m ∴ 30 × 1 cm = 30 × 1 100 m = 30 100 m = 0 , 3 m 100 cm = 1 m 1 cm = 1 100 m ∴ 30 × 1 cm = 30 × 1 100 m = 30 100 m = 0 , 3 m
      (2)
    • The applied force and distance moved are in opposite directions. Therefore, if we take s=0.3 m, then F∥=-40 N.
  2. Step 2. Analyze the question to determine what is being asked:
    • We are asked to find the work done on the car by you. We know that work done is W=F∥sW=F∥s
  3. Step 3. Substitute the values and calculate the work done:

    Again we have the applied force and the distance moved so we can proceed with calculating the work done:

    W = F ∥ s = ( - 40 N ) ( 0 . 3 m ) = - 12 J W = F ∥ s = ( - 40 N ) ( 0 . 3 m ) = - 12 J
    (3)

Note that the answer must be negative as the applied force and the motion are in opposite directions. In this case the car does work on the person trying to push.

Exercise 3: Calculating Work Done III

Calculate the work done on a box, if it is pulled 5 m along the ground by applying a force of F=10 N at an angle of 60∘60∘ to the horizontal.

Figure 1
Figure 1 (ch24_004.png)

Solution

  1. Step 1. Analyze the question to determine what information is provided :
    • The force applied is F=10 N
    • The distance moved is s=5 m along the ground
    • The angle between the applied force and the motion is 60∘60∘

    These quantities are in the correct units so we do not need to perform any unit conversions.

  2. Step 2. Analyze the question to determine what is being asked :
    • We are asked to find the work done on the box.
  3. Step 3. Calculate the component of the applied force in the direction of motion:

    Since the force and the motion are not in the same direction, we must first calculate the component of the force in the direction of the motion.

    Figure 2
    Figure 2 (ch24_005.png)

    From the force diagram we see that the component of the applied force parallel to the ground is

    F | | = F · cos ( 60 ∘ ) = 10 N · cos ( 60 ∘ ) = 5 N F | | = F · cos ( 60 ∘ ) = 10 N · cos ( 60 ∘ ) = 5 N
    (4)
  4. Step 4. Substitute and calculate the work done :

    Now we can calculate the work done on the box:

    W = F ∥ s = ( 5 N ) ( 5 m ) = 25 J W = F ∥ s = ( 5 N ) ( 5 m ) = 25 J
    (5)

Note that the answer is positive as the component of the force F∥ is in the same direction as the motion.

References

  1. Sweller, J and Cooper, G. A. (1985). The use of worked examples as a substitute for problem solving in learning algebra. Ethics and Behavior, 2(1), 59-80.

Content actions

Download module as:

PDF | EPUB (?)

What is an EPUB file?

EPUB is an electronic book format that can be read on a variety of mobile devices.

Downloading to a reading device

For detailed instructions on how to download this content's EPUB to your specific device, click the "(?)" link.

| More downloads ...

Add module to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks