Transcript lesson 1

Conservation of Energy
► Energy
 Work
► Kinetic
and Potential Energy
 Conservative and non-conservative forces
 Other forms of energy
Introduction
► Forms
of energy
 Mechanical energy
►Focus
► Forms
for now
of energy
 Energy can be transformed from one form to
another
►Essential
to the study of physics, chemistry, biology,
geology, astronomy
 Can be used in place of Newton’s law to solved
certain problems more easily.
Work
► Provides
a link between force and energy
► Work is the product of the component of
the force along the direction of the
displacement and the magnitude of the
displacement
 W=F(cosq)d
–F(cosq) is the component of the
force in the direction of the
displacement
–d is the displacement
Work
► This
gives no information about
 The time it took for the displacement to occur
 The velocity of acceleration of the object
► Note:
work is zero when
 There is no displacement (holding a bucket)
 Force and displacement are perpendicular
to each other (if we are carrying the bucket
horizontally, gravity does not work)
http://lectureonline.cl.msu.edu/~mmp/kap5/work/work.htm
More about Work
►
►
►
Work is a scalar quantity
Units of work are Nm or Joules (J)
Work can be positive or negative
 Positive if the force and the displacement
are in the same direction
 Negative if the force and the displacement
are in the opposite direction
►
Example lifting a cement block
 Work done by the person
► Is
positive when lifting the box
► Is negative when lowering the box
Examples of Work Calculations
W=F(cosq)d
Since there is no
angle
W=Fd
=(100N)5m
= 500J
W=F(cosq)d
=(100N)(cos30
)5m
= 433J
W=F(cosq)d
Since the force required to
lift up is equal and opposite
to gravity then F=+mg so
W=+mgd
W=(15kg)(9.81m/s2)5m
W= 735J
Example 4
►
A 10-N forces is applied to
push a block across a
friction free surface for a
displacement of 5.0 m to
the right.
Since Fapp is the only
horizontal force, it is the
only force that does work
•W = Fd
• = (10N)(5.0m)
• = 50J
Example 5
►
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
Since the object
moves horizontally,
only horizontally
forces will do work
Wapp = Fappd
W
= 10N 5.0 m
= 50 J
Wfrct = Ff d
= -10N(5.0 m)
= -50J
Graphing Work
►
A graph of force exerted over a displacement can be used to determine
work. Since Work = Force x displacement and Area = length x width.
If the axes on a graph are force and distance then the area under the
line will be equivalent to work done.
Find the work done over the 10 m displacement.
Area = work, there are 3 distinct areas under the line the sum will equal
total work done.
Area = ½ bh + lw + ½ bh
= ½ 3m(20N) + 5m(20N) + ½ 2m (20N)
= 30 J + 100 J + 20 J
= 150J
Work done is 150 J
Assignment 1
► Do
questions 1 – 7 in workbook