1. Conservation of Energy
Physics 11

2. Comprehension Check
A truck pushes a car by exerting a horizontal force of 500. N on it. A frictional force of 300. N opposes the car’s motion as it moves 4.0m.
A) Calculate the work done on the car by the truck.
B) Calculate the work done on the car by friction.
C) Calculate the work done on the car overall (net work).

3. Answers A) W = Fd = 500 x 4 = 2000 N B) W = Fd = -300 x 4 = -1200 J
C) Wnet = 2000 – 1200 = 8.0 x 102 J

4. Comprehension Check
Calculate the work done by a horse that exerts an applied force of 100. N on a sleigh, if the harness makes an angle of 30° with the ground and the sleigh moves 30.m across a flat, level ice surface (ie, no friction).

5. Answer
W = Fd cosΘ = (100)(30)cos(30)
= 2.6 x 103 J

6. Comprehension Check
A 50. kg crate is pulled 40. m along a horizontal floor by a constant force exerted by a person (100. N) which acts at an angle of 37°. The floor is rough and exerts a force of friction of 50.N. Determine the work done by EACH FORCE acting on the crate, and the net work done on the crate.
DRAW A DIAGRAM!!!

7. WFg = FdcosΘ  Work is 0J as the force is perpendicular to gravitational force.
WFN = 0J (same reason as above)
WFapp = Fdcos Θ =(100)(40)cos37° = 3195J  S.F.  3200 J
WFf = Fd = -50(40) = J
-2.0 x 103J
Wnet = 3200 – 2000 = 1200 J

8. Comprehension Check
Mrs. Evans is holding a 2.4kg textbook at a height of 3.4m above the floor.
a) What is the type of energy (potential or kinetic)? How do you know?
b) How much energy is there (use your equation)?
c) What is the velocity of the book at this point (ie, velocity initial)?
d) If Mrs. Evans drops the book, what is the final velocity assuming she doesn’t throw it (use your kinematics equations!)?
e) If Mrs. Evans drops the book as in d), what is the type of energy when the book hits the floor?
f) How much of this energy is there when it touches the floor?
g) Is there any time when there are both kinds of energy? If so, when? Explain.

9. Answers
A) Potential: It is not moving, it has the potential to move (fall).
B) PE = mgh = 2.4x9.81x3.4 = 80.J
C) v = 0 (at rest, not thrown)
D) vf2 = vi2 + 2ad = 2(9.81)(3.4) =
Vf = 8.2m/s [down]
KE = ½ mv2 = ½ (2.4)(8.2)2 = 80.J
80.J
When the object is falling, there is both PE and KE. When it falls 1.7m, there is equal PE and KE. Before this point (higher than 1.7m) there is more PE. After this (lower than 1.7m) there is more KE.

10. Work-Energy Theorem (copy)
“The net work done on an object is equal to its change in energy"
If the object is experiencing EK:
if Wnet is +ve, EK increases (moves in direction of force or speeds up)
if Wnet is -ve, EK decreases (moves in direction of friction or slows down)
If the object is experiencing EP:
if Wnet is +ve, EP increases (is lifted)
if Wnet is -ve, EP decreases (is lowered)

11. Total Energy and Work-Energy Theorem (Copy)
The total energy of an object is the kinetic energy added to the potential energy.
ET = Ek + Eg + Ee
As an object is dropped, the potential energy changes to kinetic energy until there is 0 EP and only EK.

12. Law of Conservation of Energy (Copy)
Within an isolated system, energy can change form, but the total amount of energy is constant
Energy cannot be created or destroyed
Ei = Ef
So Far…
Eki + Egi = Ekf + Egf

14. Mechanical Energy (copy)
The sum of kinetic and potential energy
***Not including light energy/sound energy/etc.
In the absence of friction, any external work produces a change in mechanical energy.

15. Conservation of Mechanical Energy
In the absence of friction, any external work produces a change in mechanical energy.
Why does there have to be an absence of friction?

16. Conservation of Mechanical Energy
If no work is done, then…

17. Conservation of Mechanical Energy (copy)
The total mechanical energy at any initial point in an ISOLATED system is equal to the total mechanical energy at any later point (in the absence of friction)
Energy can be transformed from one type to another
Example: A falling book starts with potential energy. As it falls, the potential energy gets transformed into kinetic energy

18. Example 1 … Think and Discuss

19. Solution
This is an example of a situation that is much easier to analyze using conservation of energy, since a kinematics analysis would involve calculating the accelerations using vectors and using the distances to calculate final velocities.
So what do we know?
Since both balls are released from rest, they have no initial kinetic energy.
Their initial potential energy is the same, since they are released from the same height.
When they reach the bottom, neither ball has any potential energy.
Same initial energy means same final energy (due to conservation of energy).
Since all of the energy is kinetic, having the same kinetic energy (and same mass) means that they are going the same speed.

20. Example 2

21. Solution

23. Example 3

24. Solution

25. SkatePark Assignment
Let’s correct!

26. Check Your Understanding
1. If the car has a speed of 12.0 m/s at point A,
What will its speed be at point C?
b) What is the highest hill (above the ground) that the car could
reach on this roller coaster?

30. Worksheets
Conservation of Mechanical Energy
Check Your Understanding

31. Check Your Understanding
1) A heavy object is dropped. If this object reaches the floor at a speed of 3.2 m/s from what height was it dropped?

32. Answer Etop = Ebottom KE + PE = KE + PE 0 + PE = KE + 0 mgh = ½ mv2
NOTE: The masses will cancel!
gh = ½ v2
9.81h = ½ (3.2)2
h = 0.52 m

33. Question 2:
A heavy box slides down a frictionless incline. The incline has a slope of 30° and the length of the incline is 12m. If the box starts from rest at the top of the incline what is the speed at the bottom?

34. Answer
V = 11 m/s

35. Question 3:
A 4.0 x 104 kg roller coaster starts from rest at point A. Assuming no friction, calculate its potential energy relative to the ground, its kinetic energy and its speed at points B,C and D in the illustration above.

36. Answer