Download Lecture07

Chapter 7: Work & Energy
THE COURSE THEME is
NEWTON’S LAWS OF MOTION!
• Chs. 4, 5 & 6: Motion analysis with forces.
• NOW (Chs. 7 & 8): An alternative analysis
using the concepts of Work & Energy.
– Easier? My opinion is yes!
• Conservation of Energy: NOT a new law!
– We’ll see that this is just Newton’s Laws of
Motion re-formulated or re-expressed (translated)
from Force Language to Energy Language.
• We’ve expressed Newton’s Laws of Motion using concepts of
position, displacement, velocity, acceleration, force.
• Newton’s Laws with Forces: Quite general (macroscopic objects).
In principle, could be used to solve any dynamics problem, But, often, they
are very difficult to apply, especially to very complicated systems. So,
alternate formulations have been developed. Often easier to apply.
• One of these is an approach that uses Energy instead of Force
as the most basic physical quantity.
• Newton’s Laws in a different language (Energy). Before we
discuss these, we need to learn vocabulary in Energy Language.
• Energy: A very common term in everyday usage. Everyday
meanings might not coincide with the PHYSICS meaning!
Every physical process involves energy or energy transfer or
transformations. Energy in physics can be somewhat abstract.
• So far, we’ve expressed Newton’s Laws of Motion in terms of
forces & we’ve considered the dynamical properties of a particle by
talking about various particle properties.
• Now, we’ll take a different approach & talk about
Systems & System Properties
• System: A small portion of the universe which we focus on in a
given problem. What the system is depends on the problem.
• A System may be, for example:
•
•
•
•
A single particle.
A collection of particles.
A region of space.
May vary in size & shape, depending on the problem
• In addition to a System, we also talk about the System Environment.
The System interacts with environment at it’s boundaries.
Sect. 7-1: Work Done by Constant Force
• Work: Precisely defined in physics. Describes what
is accomplished by a force in moving an object
through a distance.
For an object moving under
the influence of a Constant Force
F, the work done (W)  the product
of the magnitude of the displacement
(d)  the component of force parallel
to the displacement (F||).
W  F||d = Fd cosθ
d
Work Done by a Constant Force
Work
W  F||d = Fd cosθ
NOTE! Valid for a CONSTANT force ONLY!
W = F||d = Fd cosθ
• Consider a simple special case
when F & d are parallel: θ = 0, cosθ = 1

W = Fd
• Example: d = 50 m, F = 30 N
W = (30N)(50m) = 1500 N m
• Work Units: Newton - meter = Joule
1 N m = 1 Joule = 1 J
W = F||d = Fd cosθ
• A force can be exerted &
do no work! Could have
d=0
W=0
Could have F  d
 θ = 90º, cosθ = 0
W=0
Example, walking at constant
velocity v with a grocery bag:
W = F||d = Fd cosθ
A Typical Problem
An object displaced by force F
on a frictionless, horizontal surface.
Free body diagram is shown 
Normal force FN & weight mg do no
work in the process, since both
are perpendicular to the displacement.
Angles for forces:
Normal force: θ = 90°, cosθ = 0
Weight: θ = 270 (or - 90°), cosθ = 0
FN
d
W = F||d = Fd cosθ
Note:
• W is a scalar (in contrast to forces, which are vectors).
• However, W can have either a positive or a negative
sign, since cosθ can be positive or negative.
• IMPORTANT: Work (as we’ll see) is a
Transfer of Energy
The System either gains energy (if W > 0)
or loses energy (if W < 0)
Example 7-1
W = F||d = Fd cosθ
m = 50 kg, FP = 100 N, Ffr = 50 N, θ = 37º
Note: The old text had this friction force!!
Your text has taken the friction away.
Note: d is the same as x in this figure!
Example (from another text)
A man cleaning a floor pulls a vacuum cleaner with a force F = 50 N at an
angle θ = 30º with the horizontal. Calculate the work done by the force F as
the vacuum cleaner is moved a distance d = 3.0 m to the right.
F = 50 N
W = Fd cosθ
= (50)(3)cos(30º) = 130 J
FN
θ = 30º
d = 3.0 m
Solving Work Problems
1. Sketch 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. Find the net work by either
a. Find the net force & then find the work it does, or
b. Find the work done by each force & add.
Example 7-2: Work on a backpack
(a) Calculate the work a hiker must do on a backpack of
mass m = 15 kg in order to carry it up a hill of height h =
10 m, as shown.
(b) Calculate the work done by gravity on the backpack.
(c) Calculate the net work done on the backpack.
For simplicity, assume that the motion is smooth & at
constant velocity (zero acceleration).
For the hiker, ∑Fy = 0 = FH - mg
 FH = mg
WH = FHdcosθ = FHh
Conceptual Example 7-3: Does the Earth do work on the Moon?
The Moon revolves around the
Earth in a nearly circular orbit,
with approximately constant
tangential speed, kept there by
the gravitational force exerted by
the Earth. Does gravity do
(a) positive work
(b) negative work, or
(c) no work at all on the Moon?
Sect. 7-2: Scalar Product of Two Vectors
• Pure math discussion for a short time.
• We now know that the work done by a constant force is:
W = F||d = Fd cosθ
• Note that both F & d are vectors, while W is a scalar.
• W has the mathematical form that mathematicians call the
Scalar Product.
Scalar Product of Two Vectors
If A & B are vectors, their Scalar Product
(or dot product) is defined as:
So, we can write:
Scalar Product of two vectors A & B:
•
So, the work done by a constant force,
W = Fd cosθ, can be written as
• Basic math properties of the Scalar Product:
1. Commutative:
2. Distributive: A(B + C) = AB + AC
Scalar Product of 2 Vectors Using Components
Scalar Product:
• From the discussion of vector components & unit vectors: A 3
dimensional vector V can be written as
V = Vxi + Vyj + Vzk
i,j,k are unit vectors along the x,y,z axes, 
Vx,Vy,Vz are the x,y,z components of V
Unit vectors i,j,k are dimensionless & have length 1
• From scalar product definition & because the angles between
i,j,k are all 90 [cos(90°) = 0, cos(0°) = 1] we can easily show:
ii = jj = kk = 1
ij= ik = jk = 0
•
For two 3 d vectors A & B using this notation:
A = Axi + Ayj + Azk, B = Bxi + Byj + Bzk
•
Consider the scalar product of A & B in this form:
AB = (Axi + Ayj + Azk)(Bxi + Byj + Bzk)
•
Rewriting this (long form, 9 terms!):
A B =
•
(AxBx)(i i) + (AxBy)(ij) + (AxBz)(ik) +
(AyBx)(ji) + (AyBy)(jj) + (AyBz)(jk) +
(AzBx)(ki) + (AzBy)(kj) + (AzBz)(kk)
Using ii = jj = kk = 1, ij= ik = jk = 0 gives
AB = AxBx + AyBy + AzBz
(1)
This form will be convenient sometimes!
•
Consider the case where A = B. Then, (1) becomes:
AA = AxAx + AyAy + AzAz or AA = (Ax)2 + (Ay)2 + (Az)2 (2)
or
AA = |A|2 = square of the length of
 The scalar product of a vector with itself
is the square of the length of that vector.
A
Example 7-4: Using the dot product
The force shown has magnitude FP = 20 N & makes an
angle θ = 30° to the ground. Calculate the work done by
this force, using the dot product, when the wagon is
dragged a displacement d = 100 m along the ground.