Mechanical Engineering Principles

(Dana P.) #1

14


Work, energy and power


At the end of this chapter you should be
able to:


  • define work and state its unit

  • perform simple calculations on work done

  • appreciate that the area under a force/
    distance graph gives work done

  • perform calculations on a force/distance
    graph to determine work done

  • define energy and state its unit

  • state several forms of energy

  • state the principle of conservation of
    energy and give examples of conversions

  • define and calculate efficiency of systems

  • define power and state its unit

  • understand that power=force×velocity

  • perform calculations involving power,
    work done, energy and efficiency

  • define potential energy

  • perform calculations involving potential
    energy=mgh

  • define kinetic energy

  • perform calculations involving kinetic
    energy=^12 mv^2

  • distinguish between elastic and inelastic
    collisions

  • perform calculations involving kinetic
    energy in rotation=^12 Iω^2


14.1 Work


If a body moves as a result of a force being applied
to it, the force is said to do work on the body. The
amount of work done is the product of the applied


force and the distance, i.e.

work done=force×distance moved in
the direction of the force

The unit of work is thejoule, J, which is defined as
the amount of work done when a force of 1 Newton
acts for a distance of 1 m in the direction of the
force. Thus,

1J=1Nm

If a graph is plotted of experimental values of force
(on the vertical axis) against distance moved (on
the horizontal axis) a force/distance graph or work
diagram is produced.The area under the graph
represents the work done.
For example, a constant force of 20 N used to
raise a load a height of 8 m may be represented on
a force/distance graph as shown in Figure 14.1. The
area under the graph shown shaded represents the
work done. Hence

work done=20 N×8m=160 J

20

10

04
Distance / m

8

Force / N

Figure 14.1

Similarly, a spring extended by 20 mm by a force
of 500 N may be represented by the work diagram
shown in Figure 14.2, where

work done=shaded area

=^12 ×base×height
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