Physics 1408-002
Principles of Physics
Lecture 11
– Chapter 7 –
February 17, 2009
Sung-Won Lee
Sungwon.Lee@ttu.edu
Announcement II
Announcement I
Lecture note is on the web
Handout (6 slides/page)
http://highenergy.phys.ttu.edu/~slee/1408/
*** Class attendance is strongly encouraged and will be
taken randomly. Also it will be used for extra credits.
HW Assignment #4 is placed on
MateringPHYSICS, and is due by 11:59pm
on Wendseday, 2/18
The solutions for Exam 1 is available now;
Next to my office (Sci. 117)
The grades for Exam 1 is “DONE”
Exam 1 Average = 72.3 %!
SI session by
Reginald Tuvilla
SI sessions will be at the following times
and location.
Monday 4:30 - 6:00pm - Holden Hall 106
Thursday 4:00 - 5:30pm - Holden Hall 106
Chapter 7
Work & Energy
7-1 Work Done by a Constant Force
The work done by a constant force is defined as the
distance moved multiplied by the component of the
force in the direction of displacement:
where ! is the angle between the force and
the displacement vectors
•!Work Done (?) by a Constant Force
•!Scalar Product of Two Vectors – Math.
•!Work Done by a Varying Force
•!Kinetic Energy and the Work-Energy Principle
A person pulling a crate along the floor. The work done by the force
F is W = Fd cos !, where d is the displacement.
Work!!
Work Example
!"
Fr
W = F !r cos!"
•! A force does no work (W=0) on the object if the force
does not move (#r=0) through a displacement
•! The work done by a force on a moving object is zero
when the force applied is perpendicular to
the displacement (cos900=0) of its point of application
7-1 Work Done by a Constant Force
In the SI system, the units of work are joules:
As long as this person does not
lift or lower the bag of
groceries, he is doing no work
on it. The force he exerts has
no component in the direction
of motion.
The normal force (n) and
the gravitational force (mg)
do no work on the object
cos! = cos 90° = 0
The force F does work
on the object.
W = F !r cos!"
7-1 Work Done by a Constant Force
Example 7-1: Work done on a crate.
A person pulls a 50-kg crate 40 m along a
horizontal floor by a constant force FP = 100 N,
which acts at a 37° angle as shown. The floor is
smooth and exerts no friction force. Determine
(a) the work done by each force acting on the
crate, and (b) the net work done on the crate.
The person does no work on the bag of groceries
since FP is perpendicular to the displacement d.
7-1 Work Done by a Constant Force
Solving work problems:
1.! Draw a free-body diagram.
2.! Choose a coordinate system.
3.! Apply Newton’s laws to determine any
unknown forces.
4.! Find the work done by a specific force.
5.! To find the net work, either
a)! find the net force and then find the work it
does, or
b)! find the work done by each force and add.
7-1 Work Done by a Constant
Force
Example 7-2: Work on a backpack.
(a) Determine the work a hiker
must do on a 15.0-kg backpack to
carry it up a hill of height h = 10.0
m, as shown. Determine also (b)
the work done by gravity on the
backpack, and (c) the net work
done on the backpack. For
simplicity, assume the motion is
smooth and at constant velocity
(i.e., acceleration is zero).
Scalar Product of Two Vectors
The scalar product of 2 vectors
is written as A . B
= It is also called the “dot product”
7-2 Scalar Product of Two Vectors
Definition of the scalar, or dot, product:
Therefore, we can write:
A . B = A*B*cos!
" is the angle between A and B
The scalar product is commutative
A.B = B .A
The scalar product obeys the distributive law of
multiplication
A . (B + C) = A . B + A . C
Dot Products of Unit Vectors
Using component form with A and B:
7-2 Scalar Product of Two Vectors
Example 7-4: Using the dot product.
The force shown has magnitude FP = 20 N and
makes an angle of 30° to the ground. Calculate
the work done by this force, using the dot
product, when the wagon is dragged 100 m
along the ground.
y
a=1i+2j+3k
x b=4i -5j+6k
z
a . b = 1x4 + 2x(-5) + 3x6 = 12
a . a = 1x1 +
2x2 + 3x3 = 14
b . b = 4x4 + (-5)x(-5) + 6x6 = 77
7-3 Work Done by a Varying Force
A particle acted on by a varying force.
F moves along the path shown from point a to point b
!
Clearly, F·d is not constant!
7-3 Work Done by a Varying Force
For a force that varies, the work done by a force F is
approximated equal to the sum of the areas of the
rectangles. (by dividing the distance up into small
pieces, finding the work done during each, and
adding them up).
7-3 Work Done by a Varying Force
In the limit that the pieces become infinitesimally
narrow, the work is the area under the curve:
or:
Work = Area under Fcos! curve#
7-3 Work Done by a Varying Force
Work done by a spring force:
The force exerted by a spring
is given by:
x = the position of the block
with respect to the equilibrium position (x = 0)
k = the spring constant and
measures the stiffness
of the spring
•! This is called Hooke’s Law
(a) Spring in normal (unstretched) position. (b) Spring is stretched by a person exerting
a force FP to the right. The spring pulls back with a force FS where FS = -kx. (c) Person
compresses the spring (x < 0) and the spring pushes back with a force FS = kx where FS
> 0 because x < 0.
Hooke’s Law
•! When x is positive
(spring is stretched), F = negative
•! When x is 0
(at the equilibrium position), F = 0
•! When x is negative
(spring is compressed), F = positive
•! The force exerted by the spring is
always directed opposite to the
displacement from equilibrium
•! F is called the restoring force
•! If the block is released it will oscillate
back and forth b/w –x and x
7-3 Work Done by a Varying Force
Plot of F vs. x. Work
done is equal to the
shaded area.
(a) A person pulls on a spring, stretching it 3.0 cm, which requires a
maximum force of 75 N. How much work does the person do? (b) If,
instead, the person compresses the spring 3.0 cm, how much work
does the person do?
Work done to stretch a spring a distance x equals the triangular area
under the curve F = kx. The area of a triangle is ! x base x altitude,
so W = !(x)(kx) = ! kx2.
7-4 Kinetic Energy and the Work-Energy
Principle
Energy was traditionally defined as the ability
to do work. We now know that not all forces are
able to do work; however, we are dealing in
these chapters with mechanical energy, which
does follow this definition.
7-4 Kinetic Energy and the Work-Energy Principle
–!v22 - v12 = 2a(x2-x1) = 2a#x.
–!multiply by 1/2m: 1/2mv22 - 1/2mv12 = ma#x
1/ mv 2 - 1/ mv 2 = F#x
–!But F = ma
2
2
2
1
–!1/2mv22 - 1/2mv12 = F#x = WF
•! We define Kinetic Energy
–!K2 - K1 = Wnet
–!Wnet = !K
(Work/kinetic energy theorem)
A change in kinetic energy
is result of doing work to
transfer energy into a system
7-4 Kinetic Energy and the Work-Energy
Principle
This means that the work done is equal to the change
in the kinetic energy:
7-4 Kinetic Energy and the Work-Energy Principle
Because work and kinetic energy can be equated,
they must have the same units: kinetic energy is
measured in joules. Energy can be considered as the
ability to do work:
•!If the net work is positive, the kinetic energy
increases.
•!If the net work is negative, the kinetic energy
decreases.
A moving hammer strikes a nail and comes to rest. The hammer exerts a
force F on the nail; the nail exerts a force -F on the hammer (Newton’s third
law). The work done on the nail by the hammer is positive (Wn = Fd >0). The
work done on the hammer by the nail is negative (Wh = -Fd).