CK-12-Physics - Intermediate

(Marvins-Underground-K-12) #1

6.1. Work http://www.ck12.org


6.1 Work


Objectives


The student will:



  • Understand how work is defined in physics

  • Be able to solve problems involving work


Vocabulary



  • scalar quantity

  • work:Force applied over a distance.


Introduction


In the language of physics,workis force applied over a distance. Note that the force may be slowing the object
down, speeding it up, or maintaining its velocity — any number of things. It is an amount equal to the distance the
object moves multiplied by the component of force in that direction. For example, suppose you climb a ladder. On
average, the force you apply upwards to climb a ladder is equal to your weight. The total work done then is your
weight,mg, multiplied by the height of the ladder,h, giving the equation:W=mgh.


This work is important because it is one way of measuringenergy. By climbing the ladder, you have developed
energy that can be used for other purposes.


Units of Work: The Joule


The SI unit for work and energy is the Joule. This is pronounced like “jewel”, and is named after an English
physicist, James Joule (1818-1889). James Joule was a small-town brewer whose pioneering discoveries about
work, energy, and heat led to the metric standard unit of work being named in his honor. Joule discovered the
relationship between heat and mechanical work.


One Joule is defined as one Newton of work applied through one meter of distance. It is abbreviated as a capital J.


1J=1N×1m


Vector Definition


Work can also be done where the force is applied in a different direction to the displacement. If an object has
traveled a distancedunder force~F, the work done on it will be equal todmultiplied by the component of~Falong
the object’s path. The component of the force on the path will beF dcosθ, whereθis the angle between the direction
of the force and the direction of the object’s movement. Consider three special cases: 1. A force is pushing the object
in the direction that it is going, so


θ= 0 ◦; cos 0◦=1 and that force is said to be doing positive work upon the object. 2. The force is pushing opposite the
object’s motion, so

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