Force on a spring lab…
• The force applied to a spring and the amount of
stretch in the spring are directly proportional…
F = ks
F
This relationship is known as Hooke’s Law
s
Slope = spring constant, k – describes the stiffness
of the spring
Area under line = stored Eel
½ bh = ½ (s)(F) = ½ (s)(ks) = ½ ks2
Now, let’s look at Gravitational Potential Energy in a
little more detail…
• Construct a free body diagram for a box being
lifted at a constant speed and write a force
equation.
Now, let’s look at Gravitational Potential Energy in a
little more detail…
• Construct a free body diagram for a box being
lifted at a constant speed and write a force
equation
F
Fg
Since speed is constant,
F = 0…
F – Fg = 0
F – mg = 0
F = mg
So, the lifting force is equal
to the weight of the box
throughout the entire lift.
Gravitational Potential Energy
• In the spring lab we graphed Force vs. Stretch
to inform us about stored energy. What
would a graph of Force vs. Height look like for
the case of lifting the box at constant speed?
Gravitational Potential Energy
• In the spring lab we graphed Force vs. Stretch
to inform us about stored energy. What
would a graph of Force vs. Height look like for
the case of lifting the box at constant speed?
F (N)
mg
Constant force
acting on object as
it is lifted higher
and higher.
1
2
3
4
h (m)
Gravitational Potential Energy
• What characteristic of our graph is getting
larger the higher the object gets lifted?
F (N)
mg
1
2
3
4
h (m)
Gravitational Potential Energy
• What characteristic of our graph is getting
larger the higher the object gets lifted?
F (N)
AREA = Eg
mg
Area = b x h = F x h
= mgh
1
2
3
4
h (m)
Eg = mgh
Kinetic Energy
• Consider a cart being pushed by a force across
a frictionless surface some distance x…
F
F
x
Change in Energy will correspond to a force times a distance, therefore
Ek
= F  x
= max
***Recall from studying motion
vf2 = 2ax + vi2
Ek
= m(½ vf2 - ½ vi2)
Ek
= ½ mvf2 - ½ mvi2
OR…
ax = ½ vf2 - ½ vi2
So, Ek
= ½mv2
Summarizing the Energy Equations
• Elastic Potential
• Gravitational Potential
• Kintetic
Eel = ½ ks2
Eg = mgh
Ek = ½ mv2
Using LOL diagrams to write equations
Consider our box once again being lifted by some
force external to the system…
Initial
Ek
F
Eg
Energy Flow
Eel
Final
Ek
Eg
Eel
Box
Earth
Working
Ek + Eg + Eel + W =
W =
Ek + Eg + Eel
mghf
Sample Problem
A moving cart hits a spring, traveling at 5.0 m/s at the time of
contact. At the instant the cart is completely stopped, by how
much is the spring compressed? (ignore friction)
Initial
EK Eg Eel
0
Energy Flow
Diagram
EK
0
Final
Eg Eel Ediss
A closer look at Working
• Something is working if it transfers energy
into or out of a system… W = E
• W is positive when energy is added to a system
• W is negative when energy is removed from a
system
Initial
EK Eg Eel
Energy Flow
Diagram
EK
0
0
+W
-W
Final
Eg Eel Ediss
Working
• In every case, when something is working, a force is
being exerted on an object as the object moves.
• Only forces acting parallel to the motion of the
object are Forces acting in a direction parallel to
displacement in order for energy transfer to occur.
FN
F
F
x
W = F  x
f
F
Fg
***In this example, FN and Fg would
not change the energy of the system, F
would add energy, f woul remove
energy
Sample Problem
A person pushes a 1500 kg car from rest with a force of 500 N.
If there was no friction, how fast would the car be moving after
pushing for 10 m?
Initial
EK Eg Eel
0
Energy Flow
Diagram
EK
0
Final
Eg Eel Ediss
Sample Problem
A person pushes a 1500 kg car from rest with a force of 500 N.
If the coefficient of friction between the car and road is 0.20,
how fast would the car be moving after pushing for 10 m?
Initial
EK Eg Eel
0
Energy Flow
Diagram
EK
0
Final
Eg Eel Ediss
Sample Problem
• A 30 g bullet is fired into ballistic gel. If the gel
exerts a force of 50000 N to stop the bullet in a
distance of 0.10 m, how fast was the bullet
moving?
Initial
EK Eg Eel
Initial
Energy Flow
Diagram
EK
Final
0
0
Final
Eg Eel Ediss