1. Chapter 7: Energy of a System

2. NEWTON’S LAWS OF MOTION!
THE COURSE THEME is
NEWTON’S LAWS OF MOTION!
Chs. 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.

3. Newton’s Laws in Energy Language.
Up to now, we’ve expressed Newton’s Laws of Motion using the concepts of position, displacement, velocity, acceleration & force.
Newton’s Laws with Forces are quite general & work well for describing the dynamics of 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 which are often easier to apply.
One of these is an approach that uses ENERGY instead of Force as the most basic physical quantity.
The discussion of Work & Energy in Chs. 7 & 8 is actually just a statement of
Newton’s Laws in Energy Language.
Before we discuss this, we need to learn some Energy Language vocabulary.

4. Systems & System Properties
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.
In our discussions of 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

5. Sect. 7.1: Systems & Environments
So far, we’ve expressed Newton’s Laws of Motion in terms of forces & we’ve considered the dynamics properties of a particle by talking about various particle properties.
Now, we 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:
Single particle.
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. System interacts with environment at it’s boundaries.

6. Sect. 7.2: Work Done by a Constant Force
Work is precisely defined in physics. It describes what is accomplished by a force in moving an object through a distance.
For an object moving under a Constant Force, the work done (W) is defined as the product of magnitude of the displacement (Δr) & the component of the force parallel to the displacement (F||):
W  F||Δr  FΔrcosθ

7. Work Done by a Constant Force
Work: W  F||Δr  FΔr cosθ
Δr  Δr
NOTE: This form is valid for a constant force ONLY!

8. W = F||Δr = FΔr cosθ
Consider a simple special case when F & d are parallel:  θ = 0, cosθ = 1
 W = FΔr
Example: Δr = 50 m, F = 30 N
W = (30N)(50m) = 1500 N m
SI Work Units:
Newton - meter  Joule
1 N m = 1 Joule = 1 J

9. Work: W  F||Δr  FΔr cosθ
Its possible to exert a force & do no work!
Could have Δr = 0
 W = 0
Could have F  Δr
 θ = 90º, cosθ = 0
Example, walking at
constant v with a
grocery bag:

10. W  F||Δr  FΔr cosθ θ = 90°, cosθ = 0 θ = 270 (or - 90°), cosθ = 0
An object is displaced by a force F on a frictionless,
horizontal surface. The free body diagram is show here
The normal force n & the weight mg do no work in the
process, since both are perpendicular to the displacement.
For the Normal Force, n,
θ = 90°, cosθ = 0
For the Weight mg,
θ = 270 (or - 90°), cosθ = 0

11. Note W = F||Δr = FΔr cosθ The System either gains energy (if W > 0)
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 (W < 0).

12. Example 7.1

13. Example m = 50 kg, FP = 100 N, Ffr = 50 N, θ = 37º W = F||Δr =FΔr cosθ