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Introduction

Imagine a vendor carrying a basket of vegetables on her head. Is she doing any work? One would definitely say yes! However, in Physics she is not doing any work! Again, imagine a boy pushing against a wall? Is he doing any work? We can see that his muscles are contracting and expanding. He may even be sweating. But in Physics, he is not doing any work!

If the vendor is carrying a very heavy load for a long distance, we would say she has lot of energy. By this, we mean that she has a lot of stamina. If a car can travel very fast, we describe the car as powerful. So, there is a link between power and speed. However, power means something different in Physics. This chapter describes the links between work, energy and power and what these mean in Physics.

You will learn that work and energy are closely related. You shall see that the energy of an object is its capacity to do work and doing work is the process of transferring energy from one object or form to another. In other words,

  • an object with lots of energy can do lots of work.
  • when work is done, energy is lost by the object doing work and gained by the object on which the work is done.

Lifting objects or throwing them requires that you do work on them. Even making electricity flow requires that something do work. Something must have energy and transfer it through doing work to make things happen.

Work

Definition 1: Work

When a force exerted on an object causes it to move, work is done on the object (except if the force and displacement are at right angles to each other).

This means that in order for work to be done, an object must be moved a distance dd by a force FF, such that there is some non-zero component of the force in the direction of the displacement. Work is calculated as:

W = F · Δ x cos θ . W = F · Δ x cos θ .
(1)

where FF is the applied force, ΔxΔx is the displacement of the object and θθ is the angle between the applied force and the direction of motion.

Figure 1: The force FF causes the object to be displaced by ΔxΔx at angle θθ.
Figure 1 (PG12C3_001.png)

It is very important to note that for work to be done there must be a component of the applied force in the direction of motion. Forces perpendicular to the direction of motion do no work.

For example work is done on the object in Figure 2,

Figure 2: (a) The force FF causes the object to be displaced by ΔxΔx in the same direction as the force. θ=0θ=0 and cosθ=1cosθ=1. Work is done in this situation. (b) A force FF is applied to the object. The object is displaced by ΔyΔy at right angles to the force. θ=90θ=90 and cosθ=0cosθ=0. Work is not done in this situation.
Figure 2 (PG12C3_002.png)

Investigation : Is work done?

Decide whether on not work is done in the following situations. Remember that for work to be done a force must be applied in the direction of motion and there must be a displacement. Give reasons for your answer.

  1. Max pushes against a wall and becomes tired.
  2. A book falls off a table and free falls to the ground.
  3. A rocket accelerates through space.
  4. A waiter holds a tray full of meals above his head with one arm and carries it straight across the room at constant speed. (Careful! This is a tricky question.)

Tip:

The Meaning of θθ: The angle θθ is the angle between the force vector and the displacement vector. In the following situations, θ=0θ=0.

Figure 3
Figure 3 (PG12C3_003.png)

As with all physical quantities, work must have units. Following from the definition, work is measured in N··m. The name given to this combination of S.I. units is the joule (symbol J).

Definition 2: Joule

1 joule is the work done when an object is moved 1 m under the application of a force of 1 N in the direction of motion.

The work done by an object can be positive or negative. Since force (FF) and displacement (ss) are both vectors, the result of the above equation depends on their directions:

  • If FF acts in the same direction as the motion then positive work is being done. In this case the object on which the force is applied gains energy.
  • If the direction of motion and FF are opposite, then negative work is being done. This means that energy is transferred in the opposite direction. For example, if you try to push a car uphill by applying a force up the slope and instead the car rolls down the hill you are doing negative work on the car. Alternatively, the car is doing positive work on you!

Tip:

The everyday use of the word "work" differs from the physics use. In physics, only the component of the applied force that is parallel to the motion does work on an object. So, for example, a person holding up a heavy book does no work on the book.

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. Analyse the question to determine what information is provided :
    • The force applied is FF=15 N.
    • The distance moved is ss=20 m.
    • The applied force and distance moved are in the same direction. Therefore, FF=15 N.

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

  2. Step 2. Analyse 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=FsW=Fs
  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
    (2)

    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

  1. Step 1. Analyse the question to determine what information is provided :
    • The force applied is FF=40 N
    • The distance moved is ss=30 cm. This is expressed in the wrong units so we must convert to the proper S.I. units (meters):
      100 cm =1m1 cm =1100m30×1 cm =30×1100m=30100m=0,3m100 cm =1m1 cm =1100m30×1 cm =30×1100m=30100m=0,3m
      (3)
    • The applied force and distance moved are in opposite directions. Therefore, if we take ss=0.3 m, then FF=-40 N.
  2. Step 2. Analyse 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=FsW=Fs
  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
    (4)

    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.

What happens when the applied force and the motion are not parallel? If there is an angle between the direction of motion and the applied force then to determine the work done we have to calculate the component of the applied force parallel to the direction of motion. Note that this means a force perpendicular to the direction of motion can do no work.

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 FF=10 N at an angle of 6060 to the horizontal.

Figure 4
Figure 4 (PG12C3_004.png)

Solution

  1. Step 1. Analyse the question to determine what information is provided :
    • The force applied is FF=10 N
    • The distance moved is ss=5 m along the ground
    • The angle between the applied force and the motion is 6060

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

  2. Step 2. Analyse 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 5
    Figure 5 (PG12C3_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
    (5)
  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
    (6)

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

Work

  1. A 10 N force is applied to push a block across a friction free surface for a displacement of 5.0 m to the right. The block has a weight of 20 N. Determine the work done by the following forces: normal force, weight, applied force.
    Figure 6
    Figure 6 (PG12C3_006.png)
  2. A 10 N frictional force slows a moving block to a stop after a displacement of 5.0 m to the right. The block has a weight of 20 N. Determine the work done by the following forces: normal force, weight, frictional force.
    Figure 7
    Figure 7 (PG12C3_007.png)
  3. A 10 N force is applied to push a block across a frictional surface at constant speed for a displacement of 5.0 m to the right. The block has a weight of 20 N and the frictional force is 10 N. Determine the work done by the following forces: normal force, weight, frictional force.
    Figure 8
    Figure 8 (PG12C3_008.png)
  4. A 20 N object is sliding at constant speed across a friction free surface for a displacement of 5 m to the right. Determine if there is any work done.
    Figure 9
    Figure 9 (PG12C3_009.png)
  5. A 20 N object is pulled upward at constant speed by a 20 N force for a vertical displacement of 5 m. Determine if there is any work done.
    Figure 10
    Figure 10 (PG12C3_010.png)
  6. Before beginning its descent, a roller coaster is always pulled up the first hill to a high initial height. Work is done on the roller coaster to achieve this initial height. A coaster designer is considering three different incline angles of the hill at which to drag the 2 000 kg car train to the top of the 60 m high hill. In each case, the force applied to the car will be applied parallel to the hill. Her critical question is: which angle would require the least work? Analyze the data, determine the work done in each case, and answer this critical question.
    Table 1
    Angle of InclineApplied ForceDistanceWork
    351.1×104N1.1×104N100 m 
    451.3×104N1.3×104N90 m 
    551.5×104N1.5×104N80 m 
  7. Big Bertha carries a 150 N suitcase up four flights of stairs (a total height of 12 m) and then pushes it with a horizontal force of 60 N at a constant speed of 0.25 m··s-1-1 for a horizontal distance of 50 m on a frictionless surface. How much work does Big Bertha do on the suitcase during this entire trip?
  8. A mother pushes down on a pram with a force of 50 N at an angle of 30. The pram is moving on a frictionless surface. If the mother pushes the pram for a horizontal distance of 30 m, how much does she do on the pram?
    Figure 11
    Figure 11 (PG12C3_011.png)
  9. How much work is done by an applied force to raise a 2 000 N lift 5 floors vertically at a constant speed? Each floor is 5 m high.
  10. A student with a mass of 60 kg runs up three flights of stairs in 15 s, covering a vertical distance of 10 m. Determine the amount of work done by the student to elevate her body to this height. Assume that her speed is constant.

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