Physics 1A
Lecture 5A
"Energy is eternal delight.”
--William Blake
What is energy?
Energy
What are some different forms of energy?
Kinetic
Rotational
Thermal
Electrical
Chemical (Molecular Bonds)
Nuclear
Gravitational
Mass energy (ΔE = Δmc2)
Magnetic
Chemical Reactions (Gibbs Free Energy)
Energy
The problem with forces is that vectors can be
difficult.
It is far more preferable to work with scalar
values that do not have direction.
So we turn to energy (a scalar) to solve problems.
Energy is an internal property of an object.
Energy is quite possibly the most universal and
widely applicable concept in science!
Energy
Energy can be transferred from one object to
another (or from one type to another).
A “closed system” is one where the total in-flow of
energy equals the total out-flow. This is the
principle of conservation of energy.
The first type of energy we will discuss is kinetic
energy, KE.
Kinetic energy quantifies the amount of motion an
object has.
The more velocity a certain object has the more
kinetic energy it has.
Energy
For a macroscopic object:
KE = (1/2)mv2
If v = 0, then KE = 0 (Lowest possible value of KE).
The units of KE are given by:
Usually we are interested in how much kinetic
energy changes between two instances of time:
Energy
Kinetic energy is a state variable.
ΔKE only depends on the initial and final state of
the system and not what happened in between.
For example, suppose a 1,000kg truck moves at
10m/s then slows down to rest at a stop light in
about 10s. What is the change in KE?
<- negative means KE lost
Energy
Was time really a factor in this calculation?
If the car had to stop instantaneously it still will
be: ΔKE = –50,000J.
This is the one flaw of the energy equations. Time
usually does not play a role.
Energy equations are useful when comparing two
states of a system. They are generally not useful
for describing how to get from one state to
another.
In class Question
Two balls are dropped from the same height from the
roof of a building. One ball has twice the mass as
the other. Air resistance is negligible for this
question. Just before hitting the ground, the heavier
ball has:
A) one-quarter the kinetic energy of the lighter ball.
B) one-half the kinetic energy of the lighter ball.
C) the same kinetic energy as the lighter ball.
D) twice the kinetic energy of the lighter ball.
E) four times the kinetic energy of the lighter ball.
Kinetic Energy
Since both balls are dropped from the same height,
they will both attain the same velocity when they
reach the ground. (v2 = vo2 + 2aΔx)
So, vheavy = vlight = v at the bottom.
But the heavier ball is twice the mass of the lighter
ball. mheavy = 2mlight
So, KElight = (1/2)mlightv2
And, KEheavy = (1/2)mheavyv2 = (1/2)(2mlight)v2
KEheavy = 2 KElight Work
Work, W, is the transfer of energy from one
object to another (or one system to another).
Work is a scalar; it is measured in Joules. If
object 1 performs work on object 2. Then, object
1 performed negative work. Object 1 lost energy.
Object 2 had positive work performed on it.
Object 2 gained energy.
The work-energy theorem tells us how work
changes energy:
Work
A force can perform work on an object.
This is the connection between forces and energy.
Let’s assume a constant force F acts on a rolling ball
in a trough at an angle θ over displacement Δx
+y
F
+x
θ
vox
Δx
Looking at the motion in the x-direction:
vx
Work
Turning to the kinematics equations:
We want to link motion and force, so solve for ax:
Plugging back into 2nd Law (only one force acting along
x-direction):
Work
Multiplying by displacement gives:
Since this second part is just ΔKE:
We can re-write the
equation by examining the
total force (not just the x
component):
Work
The work performed on an object is equal to the
displacement times the force parallel to that
displacement.
An object must travel at least a small distance to
have work performed on it.
The sign convention for work is as follows:
If θ < 90, then cos(θ) will be positive => positive W.
If θ > 90, then cos(θ) will be negative => negative W.
If θ = 90, then cos(θ) will be zero => zero W.
Kinetic Energy
Example
A 2.00kg ball is thrown upward with a velocity
of 20.0m/s. What is the work done by gravity
from the point of release of the ball until it
reaches its highest point? How high does it go?
Answer
First, you must define a coordinate system.
Let’s choose the upward direction as positive
and make the floor to be x = 0.
Kinetic Energy
Answer
+x
Δx
<- angle between is 180o
Fg
Next, use the work-energy theorem:
At its highest point, vf = 0
this means:
Kinetic Energy
Answer
Now to find the maximum height turn to the
relationship between work and force:
Note: This is another way to find height other
than the kinematics equations.
For Next Time (FNT)
Keep Working on HW for Chapter 5.
Finish Reading Chapter 5.