FHSST Physics/Work and Energy/Work

From Wikibooks, open books for an open world
< FHSST Physics‎ | Work and Energy
Jump to: navigation, search
The Free High School Science Texts: A Textbook for High School Students Studying Physics
Main Page - << Previous Chapter (Momentum) - Next Chapter (Collisions and Explosions) >>
Work and Energy
Definition - Work - Energy - Mechanical Energy and Energy Conservation - Important Quantities, Equations, and Concepts - Sources

Work[edit]

To do work on an object, one must move the object by applying a force with at least a component in the direction of motion. The work done is given by

W= F_{\|} s
W  : work done (N.m or J)
F|  : component of applied force parallel to motion (N)
s  : displacement of the object (m)


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.

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

Definition:

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


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

  • If F| 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 F| 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!

Worked Example 38 Calculating Work Done I[edit]

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

Answer:

Step 1 : Analyse the question to determine what information is provided

  • The force applied is F = 15N.
  • The distance moved is s = 20m.
  • The applied force and distance moved are in the same direction. Therefore, F_{\|}=15N

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


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 = F|s

Step 3 : Next we substitute the values and calculate the work done

\begin{matrix}W&=&F_{\|} s\\&=& (15N)(20m)\\&=& 300\ N\cdot m\\&=& 300\ J\end{matrix}


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.

Worked Example 39 Calculating Work Done II[edit]

Question: 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 40N directed up the slope, but it slides downhill 30cm?

Answer:

Step 1 : Analyse the question to determine what information is provided

  • The force applied is F = 40N
  • The applied force and distance moved are in opposite directions. Therefore, if we take s = 0.3m, then F_{\|}=-40N.

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 = F|s

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:

\begin{matrix}W&=&F_{\|} s\\&=& (- 40N)(0.3m)\\&=& -12N\cdot m\\&=& -12\ J\end{matrix}

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.

Worked Example 40 Calculating Work Done III[edit]

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

Answer:

Step 1 :

Analyse the question to determine what information is provided

  • The force applied is F = 10N
  • The distance moved is s = 5m along the ground
  • The angle between the applied force and the motion is 60o

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

Step 2 : Analyse the question to determine what is being asked

  • We are asked to find the work done on the box.

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.

Fhsst wrkeng1.png

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

\begin{matrix}F_{||}&=&F\cdot \cos(60^o)\\&=& 10N\cdot \cos(60^o)\\&=& 5\ N\end{matrix}

Step 4 : Substitute and calculate the work done

Now we can calculate the work done on the box:

\begin{matrix}W&=&F_{\|} s\\&=& (5N)(5m) \\&=& 25\ J\end{matrix}

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


We will now discuss energy in greater detail.