1. Internal vs. External Forces
Internal Forces
External Forces
Fgrav
Fspring
Fapp
Ffrict
Fair
Ftens
Fnorm

2. Internal Forces
When work is done by an internal force total mechanical energy (KE & PE) is conserved.
The sum of KE and PE remains Constant
3 Plicker Questions

3. External Forces
When work is done by an external force total mechanical energy (KE & PE) is changed.
If work is positive the object gains energy
Force and displacement are in the same direction
If work is negative the object loses energy
Force and displacement are in the opposite direction
Energy gain/loss can be in the form of PE, KE, or Both
“Non conservative forces”
Only if external Force has a vertical component can it contribute to PE change
5 Plicker Q

4. Relating Work to Energy
Total mechanical energy plus work done by external forces is equal to the final total mechanical energy.
TMEi + Wext = TMEf or
KEi + PEi + Wext = Kef + PEf

5. Example Problem
Consider a weightlifter who applies an upwards force (say 1000 N) to a barbell to displace it upwards a given distance (say 0.25 meters) at a constant speed. The initial energy plus the work done by the external force equals the final energy. If the barbell begins with 1500 Joules of energy (this is just a made up value) and the weightlifter does 250 Joules of work (F•d•cosine of angle = 1000 N•0.25 m•cosine 0 degrees = 250 J), then the barbell will finish with 1750 Joules of mechanical energy. The final amount of mechanical energy (1750 J) is equal to the initial amount of mechanical energy (1500 J) plus the work done by external forces (250 J).

6. Example Problem
Now consider a baseball catcher who applies a rightward force (say 6000 N) to a leftward moving baseball to bring it from a high speed to a rest position over a given distance (say 0.10 meters). The initial energy plus the work done by the external force equals the final energy. If the ball begins with 605 Joules of energy (this is just another made up value), and the catcher does -600 Joules of work (F•d•cosine of angle = 6000 N•0.10 m•cosine 180 degrees = -600 J), then the ball will finish with 5 Joules of mechanical energy. The final energy (5 J) is equal to the initial energy (605 J) plus the work done by external forces (-600 J).

7. Example Problem
As a final example, consider a cart being pulled up an inclined plane at constant speed by a student during a Physics lab. The applied force on the cart (say 18 N) is directed parallel to the incline to cause the cart to be displaced parallel to the incline for a given displacement (say 0.7 m). The initial energy plus the work done by the external force equals the final energy. If the cart begins with 0 Joules of energy (this is just another made up value), and the student does 12.6 Joules of work (F•d•cosine of angle = 18 N•0.7 m•cosine 0 degrees = 12.6 J), then the cart will finish with 12.6 Joules of mechanical energy. The final energy (12.6 J) is equal to the initial energy (0 J) plus the work done by external forces (12.6 J).

8. A 1000-kg car traveling with a speed of 25 m/s skids to a stop
A 1000-kg car traveling with a speed of 25 m/s skids to a stop. The car experiences an 8000 N force of friction. Determine the stopping distance of the car.
~39m

9. At the end of the Shock Wave roller coaster ride, the 6000-kg train of cars (includes passengers) is slowed from a speed of 20 m/s to a speed of 5 m/s over a distance of 20 meters. Determine the braking force required to slow the train of cars by this amount.

10. A shopping cart full of groceries is sitting at the top of a 2
A shopping cart full of groceries is sitting at the top of a 2.0-m hill. The cart begins to roll until it hits a stump at the bottom of the hill. Upon impact, a 0.25-kg can of peaches flies horizontally out of the shopping cart and hits a parked car with an average force of 500 N. How deep a dent is made in the car (i.e., over what distance does the 500 N force act upon the can of peaches before bringing it to a stop)?

11. Stopping Distance
Force that does work over a distance to remove mechanical energy from an object.
TMEi + Wext = TMEf
Then stopping is KEi + Wext= 0J
Then .5*m*v2 + F*d*cos180= 0J
Therefore .5*m*v2 = F*d
And v2 α d
stopping distance depends on velocity squared
If velocity doubles, distance quadruples

12. Mechanical Energy Conserved
If only internal forces are doing work then there is no change in TME, therefore:
KEi + PEi = KEf + Pef
PE increases KE decreases (& vice versa)
PE is transformed into KE (& vice versa)
TME is conserved

13. WARNING: PLEASE IGNORE AIR RESISTANCE & FRICTION
Pendulum Motion
What forces act upon a pendulum that is swinging? (hint one is internal the other is external)
Is the external force doing work?
Is energy conserved?
What happens to PE and KE as height decreases?
Where do you expect the pendulum to have the greatest PE? (A, B, or C)
Where do you expect the pendulum to have the greatest KE? (A, B, or C)
WARNING: PLEASE IGNORE AIR RESISTANCE & FRICTION

14. TME Conserved Systems
Roller Coasters
Ski Jumpers