1. Area under line = stored Eel
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

2. 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.

3. 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
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. Fg

4. 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?

5. 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.
h (m) 1 2 3 4

6. Gravitational Potential Energy
What characteristic of our graph is getting larger the higher the object gets lifted?
F (N) mg
h (m) 1 2 3 4

7. Gravitational Potential Energy
What characteristic of our graph is getting larger the higher the object gets lifted?
F (N)
AREA = Eg
Area = b x h = F x h
= mgh
Eg = mgh mg
h (m) 1 2 3 4

8. 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 + vi OR…
ax = ½ vf2 - ½ vi2
Ek = m(½ vf2 - ½ vi2)
Ek = ½ mvf2 - ½ mvi2 So, Ek = ½mv2

9. Summarizing the Energy Equations
Elastic Potential Eel = ½ ks2
Gravitational Potential Eg = mgh
Kintetic Ek = ½ mv2

10. Using LOL diagrams to write equations
Consider our box once again being lifted by some force external to the system…
Initial Energy Flow Final
Ek Eg Eel Ek Eg Eel F
Box
Earth
Working
Ek + Eg + Eel + W = Ek + Eg + Eel
W = mghf

11. 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)

12. 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
+W - W

13. 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.
W = F  x FN F F f F Fg x
***In this example, FN and Fg would not change the energy of the system, F would add energy, f woul remove energy

14. 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?

15. 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?

16. Sample Problem
A 30 g bullet is fired into ballistic gel. If the gel exerts a force of N to stop the bullet in a distance of 0.10 m, how fast was the bullet moving?
Energy Flow
Diagram EK
Eel
Initial Eg
Final
Ediss
Initial
Final