1. The meat & potatoes of physics AP Physics Part 1
Work & energy
The meat & potatoes of physics
AP Physics Part 1

2. Work
Work has a very specific definition in physics: work is when a force causes an object to be moved
The original definition was “weight lifted through a height”
In a “perfect” case, the work done on a system would equal the energy change in the system… but watch out for friction
Work is measured in Joules:
W = FDr
A Joule is also a Newton-meter.

3. Work
In a simple case of lifting a weight, the relationship is direct: the work done on the weight equals its change in potential energy.
If you push on something and it doesn’t move, then no matter how much you tire yourself out you have done no work.

4. Work
Force
If you’re just holding or carrying a box around, your arms aren’t doing any work in the physics sense
The force you’re exerting is perpendicular to the direction you’re moving – it doesn’t do work
Work is the ability to create a meaningful change, like lifting something.
Direction of Movement

5. Problems
You hold a box that weighs 60N and walk 10m across the room. Your arms are aching… How much work did you just do?
Knowns Unknowns Equations Solve it!
Force = 60N
Distance = 10m
W = ?
W = Fd
All that force wasn’t in the direction of movement, so:
W = 0

6. Problems
You pick up a 6 kg (60N) box and lift it steadily to a height of 2m. How much work did you do?
Knowns Unknowns Equations Solve it!
Force = 60N
Distance = 2m
W = ?
W = Fd
PE = mgh
Solve it either way:
W = Fd = 60N * 2m
W = 120J
PE = mgh = 6kg * 10 * 2m
PE = 120J… the amount of work you did on the box

7. Work
What is the force is at an angle q to the direction of displacement?
We use the component of the force which is in the direction of the displacement:
W = F•Dr = FDrcosq

8. Problems
A force is applied to a crate at an angle of 25°. The crate is dragged across a deck a distance of 2.5m. If the amount of work that was done was 1,210J, what was the force?
W = FDrcosq
F = W/Drcosq
F = 1,210J/((2.5m)(cos25°)) = N

9. Work
Work is a scalar. No need to include direction in any description of work done.
Joules are a small unit, so we often deal in kJ or MJ. A Joule is about the work you do lifting a Big Mac one meter.

10. Energy
In physics terms, energy is the capacity to do work — it’s stored work. Something with energy has work in it.
Energy comes in many forms: kinetic, potential, electrical, solar, nuclear, chemical…and it’s all measured in Joules.
It’s really all down to potential and kinetic.

11. Kinetic energy
The energy of motion
Measured in Joules
K = ½ mv2

12. Problems
Usain Bolt masses 94 kg and runs at 10 m/s. How much kinetic energy does he have?
Knowns Unknowns Equations Solve it!
Mass = 94kg
Speed = 10m/s
K = ?
K = ½ mv2
K = 94 * (10m/s)2 /2
K = 4,700J

13. Problems
Usain’s brother, Insane Bolt, runs twice as fast. How much kinetic energy does he have?
Knowns Unknowns Equations Solve it!
Mass = 94kg
Speed = 20m/s
K = ?
K = ½ mv2
K = 94 * (20m/s)2 /2
K = 18,800J
Twice as fast means 4x as much kinetic energy.
4,700x4=18,800J

14. Problems
A net 6,500N force is applied to a 1500kg car, moving it forward. What is its kinetic energy and speed after it has been displaced 150m?
W = FDr = 6,500N(150m) = 975,000J
That’s how much kinetic energy it has, so K = 975,000J
K = ½ mv2 … so v = √(2K/m)
v = m/s

15. Potential energy Ug = mgh
Energy waiting to be put into motion, or into use
Stored energy
Includes gravitational and elastic
ALSO measured in Joules
Ug = mgh

16. Potential energy
Potential energy is relative! What matters is how far something has been lifted or how far it has dropped.
It is based on position or condition.
Wnet = DU = U – U0
Choose a reference frame that simplifies the problem – set the lowest level to zero and set other heights above that.

17. Problems
How much potential energy does a 40kg rock have at top of a 10m cliff?
Knowns Unknowns Equations Solve it!
Mass = 40kg
Height = 10m
PE = ?
PE = mgh
PE = 40kg * 10 * 10m
PE = 4,000J

18. Conservation of Energy
Energy is neither drained nor lost in any process.
The energy you start with equals the energy you end with
Energy can be transformed from one form into another
Energy isn’t really lost, but it can be wasted by being transformed into heat

19. Worksheet 1
Use pie charts to analyze the energy changes in the situations given.
Label the pies to correspond with the positions of the objects as given the pictures
The pies should be accurately divided and labeled with the energy storage mechanisms involved.
When energy is dissipated, you can show it as a slice of the pie, as a shrinking pie OR as a dark rim around the outside of the pie. Don’t add dissipated energy until #4.

20. Popper Activity
Using the concept of energy transformation and conservation, determine
The amount and type of energy in the popper before it pops
The amount and type of energy just after the popper pops
The velocity of the popper just after it pops
The amount and type of energy when the popper reaches its highest point
The amount and type of energy just before the popper lands—and its velocity at that time
The spring constant for the popper—measure the x which is the invert distance.
Measure at least 5 good pops—keep a table.
Show your calculations.
TE=total energy

21. Conservation of Energy
Energy can change back and forth between several types in a system

22. Conservation of Energy
A pendulum is a good example of this – energy changes back and forth between kinetic and potential

23. Conservation of Energy
This allows us to solve very complicated problems… if we ignore friction.

24. Conservation of Energy
In any situation, total energy before equals total energy after.
E = E’
In a situation where an object might have both kinetic and potential energy before and after:
K + Ug = K’ + Ug’
½ mv02 + mgh0 = ½ mv2 + mgh’

25. Problems ½ mv02 + mgh0 = ½ mv2 + mgh’ mgh0 = ½ mv2
A 1.5kg ball is dropped from a height of 2.3m. What is its velocity just before it hits the deck?
½ mv02 + mgh0 = ½ mv2 + mgh’
It has no initial kinetic energy and no final potential energy…
mgh0 = ½ mv2
So final kinetic energy equals starting potential energy.
½ mv2 = mgh v = √(2gh) = √(2x9.8x2.3)
v = m/s

26. Problems ½ mv02 + mgh0 = ½ mv2 + mgh’ mgh0 = ½ mv2
A roller coaster pauses at top of a 75m hill. What will its velocity be at the bottom of the hill?
½ mv02 + mgh0 = ½ mv2 + mgh’
It has no initial kinetic energy and no final potential energy…
mgh0 = ½ mv2
So final kinetic energy equals starting potential energy.
½ mv2 = mgh v = √(2gh) = √ (2x9.8x75)
v = m/s

27. Problems ½ mv02 + mgh0 = ½ mv2 + mgh’ mgh0 = ½ mv2 + mgh
A roller coaster pauses at top of a 94m hill. The next hill is 68m tall. The roller rolls down the first hill and up the second – how fast is it going at the top?
½ mv02 + mgh0 = ½ mv2 + mgh’
It has no initial kinetic energy but it does have final potential energy…
mgh0 = ½ mv2 + mgh
½ mv2 = mgh0 - mgh v = √(2g(h0 – h))
v = √(2x9.8(94-68))
v = m/s

28. Problems
A sled and rider together have a mass of 87kg. They are at top of a hill which is at an angle of 42.5°. They slide down the hill to the bottom, a distance of 35m. How fast are they going at the bottom?
You COULD figure out the angles, components of gravitational acceleration, etc… but it’s easier to solve this with energy.
mgy = ½ mv2
We need to calculate the height. What is y?
v = √(2gy) … which is (2x9.8x35sin42.5°)

29. Problems y = dsinq v = √(2gy) … which is (2x9.8x35sin42.5°)
A sled and rider together have a mass of 87kg. They are at top of a hill which is at an angle of 42.5°. They slide down the hill to the bottom, a distance of 35m. How fast are they going at the bottom?
y = dsinq
v = √(2gy) … which is (2x9.8x35sin42.5°)
v = 21.52m/s