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Work, Energy, and Energy Resources

42 Work: The Scientific Definition

Learning Objectives

  • Explain how an object must be displaced for a force on it to do work.

  • Explain how the relative directions of force and displacement determine whether work done is positive, negative, or zero

 

The Scientific Definition of Work

In everyday life, we often describe many activities as “work”—such as writing an exam, holding a heavy object, or walking with groceries. But in physics, the word work has a more precise and limited meaning.

In physics, work is done only when a force causes displacement in the direction of that force.

If there is no movement, or if the movement is perpendicular to the applied force, then no work (in the scientific sense) is done on the object.

The Work Formula

The formal definition of work involves both force and displacement, as well as the angle between them. When a constant force [latex]\mathbf{F}[/latex] is applied and an object is displaced by a distance [latex]\mathbf{d}[/latex], the work done is:

[latex]W = |\mathbf{F}| \cdot |\mathbf{d}| \cdot \cos(\theta)[/latex]

or simply:

[latex]W = Fd\cos\theta[/latex]

Where:

  • [latex]W[/latex] is the work done by the force (in joules, J),

  • [latex]F[/latex] is the magnitude of the force (in newtons, N),

  • [latex]d[/latex] is the displacement of the object (in meters, m),

  • [latex]\theta[/latex] is the angle between the force vector and the displacement vector.

Interpreting the Equation

The value of [latex]\cos(\theta)[/latex] determines whether the work is:

  • Positive: when the force and displacement are in the same direction ([latex]\theta = 0^\circ[/latex]),

  • Negative: when the force and displacement are in opposite directions ([latex]\theta = 180^\circ[/latex]),

  • Zero: when the force is perpendicular to displacement ([latex]\theta = 90^\circ[/latex]).

Real-Life Examples of Work

Five drawings labeled a through e. In (a), person pushing a lawn mower with a force F. Force is represented by a vector making an angle theta with the horizontal and displacement of the mower is represented by vector d. The component of vector F along vector d is F cosine theta. Work done by the person W is equal to F d cosine theta. (b) A person is standing with a briefcase in his hand. The force F shown by a vector arrow pointing upwards starting from the handle of briefcase and the displacement d is equal to zero. (c) A person is walking holding the briefcase in his hand. Force vector F is in the vertical direction starting from the handle of briefcase and displacement vector d is in horizontal direction starting from the same point as vector F. The angle between F and d theta is equal to 90 degrees. Cosine theta is equal to zero. (d) A briefcase is shown in front of a set of stairs. A vector d starting from the first stair points along the incline of the stair and a force vector F is in vertical direction starting from the same point as vector d. The angle between them is theta. A component of vector F along vector d is F d cosine theta. (e) A briefcase is shown lowered vertically down from an electric generator. The displacement vector d points downwards and force vector F points upwards acting on the briefcase.
Figure 42.1: Examples of work. (a) The work done by the force [latex]\mathbf{F}[/latex] on this lawn mower is [latex]\text{Fd}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta [/latex]. Note that [latex]F\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta [/latex] is the component of the force in the direction of motion. (b) A person holding a briefcase does no work on it, because there is no displacement. No energy is transferred to or from the briefcase. (c) The person moving the briefcase horizontally at a constant speed does no work on it, and transfers no energy to it. (d) Work is done on the briefcase by carrying it up stairs at constant speed, because there is necessarily a component of force [latex]\mathbf{F}[/latex] in the direction of the motion. Energy is transferred to the briefcase and could in turn be used to do work. (e) When the briefcase is lowered, energy is transferred out of the briefcase and into an electric generator. Here the work done on the briefcase by the generator is negative, removing energy from the briefcase, because [latex]\mathbf{F}[/latex] and [latex]\mathbf{d}[/latex] are in opposite directions.

Let’s interpret a few common scenarios using the scientific definition of work:

Figure 42.1(a) – Pushing a lawn mower:
The force [latex]\mathbf{F}[/latex] is applied at an angle [latex]\theta[/latex] to the horizontal. Only the horizontal component of the force contributes to the work done:
[latex]W = Fd\cos\theta[/latex]

Figure 42.1(b)  – Holding a briefcase stationary:
There is a force (upward) but no displacement. Therefore,
[latex]W = 0[/latex] — no work is done on the briefcase.

Figure 42.1(c)  – Carrying a briefcase horizontally at constant speed:
The upward force from your arm is perpendicular to the forward displacement, so:
[latex]\cos(90^\circ) = 0[/latex]
Again, [latex]W = 0[/latex] — no energy is transferred to the briefcase.

Figure 42.1(d) – Carrying a briefcase upstairs:
The upward displacement is in the same direction as the force applied. Here, work is positive and energy is transferred to the briefcase–Earth system.

Figure 42.1(e)  – Lowering a briefcase into a generator:
Here, the generator applies an upward force while the displacement is downward. The angle between force and displacement is [latex]180^\circ[/latex], so:
[latex]\cos(180^\circ) = -1[/latex]
This results in negative work, meaning energy is removed from the briefcase and transferred to the generator.

Units of Work

Since work is the product of force and displacement, its SI unit is:

[latex]1~\text{joule} = 1~\text{newton} \cdot \text{meter} = 1~\text{J} = 1~\text{kg} \cdot \text{m}^2/\text{s}^2[/latex]

  • 1 joule is a small amount of energy; for example, lifting a 100-gram apple about 1 meter upward requires roughly 1 joule of work.

Example 42.1: Calculating the Work You Do to Push a Lawn Mower Across a Large Lawn

How much work is done on the lawn mower by the person in Figure 42.1(a) if he exerts a constant force of [latex]\text{75}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{N}[/latex] at an angle [latex]\text{35}\text{º}[/latex] below the horizontal and pushes the mower [latex]\text{25}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{m}[/latex] on level ground? Convert the amount of work from joules to kilocalories and compare it with this person’s average daily intake of [latex]\text{10},\text{000}\phantom{\rule{0.25em}{0ex}}\text{kJ}[/latex] (about [latex]\text{2400}\phantom{\rule{0.25em}{0ex}}\text{kcal}[/latex]) of food energy. One calorie (1 cal) of heat is the amount required to warm 1 g of water by [latex]1\text{º}\text{C}[/latex], and is equivalent to [latex]4\text{.}\text{184}\phantom{\rule{0.25em}{0ex}}\text{J}[/latex], while one food calorie (1 kcal) is equivalent to [latex]\text{4184}\phantom{\rule{0.25em}{0ex}}\text{J}[/latex].

Strategy

We can solve this problem by substituting the given values into the definition of work done on a system, stated in the equation [latex]W=\text{Fd}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta[/latex]. The force, angle, and displacement are given, so that only the work [latex]W[/latex] is unknown.

Solution

The equation for the work is

[latex]W=\text{Fd}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta .[/latex]

Substituting the known values gives

[latex]\begin{array}{lll}W& =& \left(\text{75.0 N}\right)\left(\text{25.0 m}\right)\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(\text{35.0º}\right)\\ & =& \text{1536 J}=\text{1.54}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{J.}\end{array}[/latex]

Converting the work in joules to kilocalories yields [latex]W=\left(\text{1536}\phantom{\rule{0.25em}{0ex}}\text{J}\right)\left(1\phantom{\rule{0.25em}{0ex}}\text{kcal}/\text{4184}\phantom{\rule{0.25em}{0ex}}\text{J}\right)=0\text{.}\text{367}\phantom{\rule{0.25em}{0ex}}\text{kcal}[/latex]. The ratio of the work done to the daily consumption is

[latex]\frac{W}{\text{2400}\phantom{\rule{0.25em}{0ex}}\text{kcal}}=1\text{.}\text{53}×{\text{10}}^{-4}\text{.}[/latex]

Discussion

This ratio is a tiny fraction of what the person consumes, but it is typical. Very little of the energy released in the consumption of food is used to do work. Even when we “work” all day long, less than 10% of our food energy intake is used to do work and more than 90% is converted to thermal energy or stored as chemical energy in fat.

Section Summary

    • Work is the transfer of energy that occurs when a force is applied to an object and the object is displaced in the direction of that force.

    • The work [latex]W[/latex] done by a constant force [latex]\mathbf{F}[/latex] acting on an object displaced by [latex]\mathbf{d}[/latex] at an angle [latex]\theta[/latex] is given by:

      [latex]W = Fd\cos\theta[/latex]

      Where:

      • [latex]F[/latex] is the magnitude of the force (in newtons),

      • [latex]d[/latex] is the displacement (in meters),

      • [latex]\theta[/latex] is the angle between the force and displacement vectors.

    • The SI unit for both work and energy is the joule (J):

      [latex]1~\text{J} = 1~\text{N} \cdot \text{m} = 1~\text{kg} \cdot \text{m}^2/\text{s}^2[/latex]

    • When is work zero?

      • When the object is not displaced ([latex]d = 0[/latex]).

      • When the force is perpendicular to the displacement ([latex]\theta = 90^\circ[/latex]).

    • Sign of work:

      • Positive work: Force and displacement are in the same direction ([latex]0^\circ \theta 90^\circ[/latex]).

      • Negative work: Force and displacement are in opposite directions ([latex]90^\circ \theta \leq 180^\circ[/latex]).

    Conceptual Questions

    1. Give an example of something we think of as work in everyday circumstances that is not work in the scientific sense. Is energy transferred or changed in form in your example? If so, explain how this is accomplished without doing work.
    2. Give an example of a situation in which there is a force and a displacement, but the force does no work. Explain why it does no work.
    3. Describe a situation in which a force is exerted for a long time but does no work. Explain.

    Problems & Exercises

    1. How much work does a supermarket checkout attendant do on a can of soup he pushes 0.600 m horizontally with a force of 5.00 N? Express your answer in joules and kilocalories.
    2. A 75.0-kg person climbs stairs, gaining 2.50 meters in height. Find the work done to accomplish this task.  (a) Calculate the work done on a 1500-kg elevator car by its cable to lift it 40.0 m at constant speed, assuming friction averages 100 N. (b) What is the work done on the lift by the gravitational force in this process? (c) What is the total work done on the lift?
    3. Suppose a car travels 108 km at a speed of 30.0 m/s, and uses 2.0 gal of gasoline. Only 30% of the gasoline goes into useful work by the force that keeps the car moving at constant speed despite friction. (See Table 47.1 for the energy content of gasoline.) (a) What is the magnitude of the force exerted to keep the car moving at constant speed? (b) If the required force is directly proportional to speed, how many gallons will be used to drive 108 km at a speed of 28.0 m/s?
    4. Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of [latex]\text{20}\text{.}0\text{º}[/latex] with the horizontal. (See Figure 42.2) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate and on his body to get up the ramp.
      A person is pushing a heavy crate up a ramp. The force vector F applied by the person is acting parallel to the ramp.
      Figure 42.2: A man pushes a crate up a ramp.
    5. How much work is done by the boy pulling his sister 30.0 m in a wagon as shown in Figure 42.3? Assume no friction acts on the wagon.
      A child is sitting inside a wagon and being pulled by a boy with a force F at an angle thirty degrees upward from the horizontal. F is equal to fifty newtons, the displacement vector d is horizontal in the direction of motion. The magnitude of d is thirty meters.
      Figure 42.3: The boy does work on the system of the wagon and the child when he pulls them as shown.

    6. Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a [latex]\text{60}\text{.}0\text{º}[/latex] slope at constant speed, as shown in Figure 42.4. The coefficient of friction between the sled and the snow is 0.100. (a) How much work is done by friction as the sled moves 30.0 m along the hill? (b) How much work is done by the rope on the sled in this distance? (c) What is the work done by the gravitational force on the sled? (d) What is the total work done?
    A person on a rescue sled is shown being pulled up a slope. The slope makes an angle of sixty degrees from the horizontal. The weight of the person is shown by vector w acting vertically downward. The tension in the rope depicted by vector T is along the incline in the upward direction; vector f depicting frictional force is also acting in the same direction.
    Figure 42.4: A rescue sled and victim are lowered down a steep slope.

    Glossary

    energy
    the ability to do work
    work
    the transfer of energy by a force that causes an object to be displaced; the product of the component of the force in the direction of the displacement and the magnitude of the displacement
    joule
    SI unit of work and energy, equal to one newton-meter
    definition

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