1. General Physics 101 PHYS Dr. Zyad Ahmed Tawfik
Website : zyadinaya.wordpress.com

2. Lecture No. 6
Work and Energy

3. Lecture No.6 Energy and work
Vectors

4. Energy
Symbol: E
Units:
J, Joule

5. Energy Energy Approach Energy definition Is the ability to do work.
The energy approach use to describing motion of valid system, such as a car or a human body

6. Energy
Every physical process that occurs in the Universe and every chemical reaction inside the human body involves energy, and this energy transfer or transformation from state to another state.

7. Types of Energy

8. Electrical Energy

9. Heat Energy

10. Chemical Energy
Chemical Energy is energy stored in the bonds of chemical compounds (atoms and molecules).
This type of energy available in nature, and the most important types of oil, coal, natural gas and wood

11. Electromagnetic Energy

12. 2- Work

13. Work
Work is the energy transferred to a system by a force that acts on it.
Units of Work
The unit of work is a joule (J)
1 joule = 1 Newton . 1 meter
J = N · m

14. J = N · m Units of Work The unit of work is a joule (J)
1 joule = 1 Newton . 1 meter
J = N · m

15. Work
Work = The Scalar Dot Product between Force F and Displacement d.
So that means if you apply a force on an object and it covers a displacement you have supplied ENERGY or done WORK on that object.
W = F . d

16. Calculate work done on an object

17. 1- With out angle
In this case it means that F and d MUST be parallel. To ensure that they are parallel we add the cosine on the end.
Apply FORCE
Displacement
So the equation used to calculate the work(W) in this case it:
W= F . d

18. Also with friction force
The equation used to calculate the work(W) in this case it:
W= Ff . d
This signal(-) Because in this case the friction force in the opposite direction for displacement
Friction FORCE
Displacement

19. Example 1,with out angle W = 50 · 10 = 500 J
A 50 N horizontal force is applied to an object over a distance of 10 m. Find the amount of work done.
Solution: since, W = F . d
W = 50 · 10 = 500 J
Object
50 N
10 m

20. With angle
In the figure above, we see the woman applying a force at an angle theta q.
Only the Horizontal component actually causes the box to move. and thus imparts energy to the box

21. W = F . d cos q So the force analysis into two components Fx and Fy
Where Fx = F cos q and Fy = F sin q
This vertical component (F sin q) is not doing any work on the box ( W=0) because it is perpendicular to the displacement). And also it is not parallel to the displacement.
So in this case, the work done given by
W = F . d cos q

22. Example1, with angle F sin F F cos
A 50 N sloping force is applied to an object with angle 50 degree over a distance of 10 m. Find the amount of work done.
Solution: since, W = F . d cos 
W = 50 · 10 cos 50 = 500 x = J
F sin F 
F cos
Object x

23. W= 50 . 5 cos30 W= 216.5 J Example2, with angle
Suppose the woman in the figure above applies a 50 N force to a box at an angle of 30 degrees above the horizontal. She manages to pull the box 5 meters.
Calculate the WORK done by the woman on the box?
Solution: since, W = F . d cos 
W = F . d cos 
W= cos30
W= J

24. 3- Kinetic Energy
and
Work
Vectors

25. K = ½ mv2 (1) What is Kinetic Energy?
Kinetic Energy is the energy of a particle due to its motion
K = ½ mv (1)
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle

26. Work and Kinetic Energy
Consider an object of mass m subjected to a constant force F , the object moves a distance x parallel to F. show figure
Let, v0 is the initial velocity and Vf is the final velocity.

27. Work and Kinetic Energy
Since, vf2 = v a d
Multiplying by m/2, this becomes
½ mvf2 = ½ m v02 + m a d
But F = m a, Also W = F d = m a d
Then, ½ mvf2 = ½ m v02 + W
But the low of Kinetic Energy(K= ½ mv2 )
so , Kf = K0 +W (1)

28. Work and Kinetic Energy
Kf = K0 + W _______(1)
Where
Kf is the final kinetic energy
K0 is the initial kinetic energy
The equation (1) say
(The final kinetic energy of an object = initial kinetic energy + total work done.
Note-1, work and kinetic energy have the same dimensions and units

29. Work and Kinetic Energy
Also from equation (1) we can Wright
W = Kf - K0 _______(2)
The equation (3) say
(total work done= The final kinetic energy of an object - initial kinetic energy)
_____________________________________________
But the equation 2 can be written in this format
W =½ mvf2 - ½ m v (3)
where k= ½ m v2
But DK = ½ mvf2 - ½ mvo2
so w= DK (5)
The equation (5) say
“the net WORK done on an object is equal to the change in kinetic energy of the object."

30. Example1,
Suppose the woman in the figure above applies a 50 N force to 25-kg a box at an angle of 30 degrees above the horizontal. She manages to pull the box 5 meters.
Calculate the WORK done by the woman on the box
The speed of the box after 5 meters if the box started from rest.
Solution:
a) W = F . d cos  so W= cos W= J
____________________________________________________
b) w= DK = ½ mv where m=25 and w=216.5
= 1/ v2
v= 4.16 m/s

31. 4- Potential
Energy
Vectors

32. Potential Energy potential energy is energy stored in an object.
Gravity gives potential energy to an object.  
This potential energy is a result of gravity pulling downwards.
Potential Energy means the work done by gravity on the object.
The potential energy of the object is depend of the mass of the object (m), gravitational acceleration of the earth (g=9.8 m/sec2) and objective height above earth's surface (h).
The formula for potential energy (U) due to gravity is U = m.g.h
P.E. = mass x height x gravity

33. Example 1 What is the P.E. of a 50-kg person at a height of 480 m?
Solution
U = m.g.h
M=50kg , g=10 and h= 480
U = m.g.h = (50 kg)(10 m/s2)(480 m)= J
U=240 K J

34. Example 2
Find the potential energy of 20 Kg mass child sitting on a roof 10 m above the ground.?
Solution
U = m.g.h
M=20kg , g=10 and h= 10
U = m.g.h = (20 kg)(10 m/s2)(10 m)=1960 J
U=2000 K J

35. Potential Energy and Kinetic energy
Total mechanical energy = kinetic energy + potential energy
E = K + u
Potential Energy and Kinetic energy

36. 5-law of conservation of mechanical energy
Vectors

37. The law of conservation of mechanical energy
The law of conservation of mechanical energy states: Energy cannot be created or destroyed, only transformed!
Energy Before
Energy After
Am I moving? If yes, Ko
Am I above the ground? If yes, Uo
Am I moving? If yes, K
Am I above the ground? If yes, U

38. Conservation of Energy
In Figure A, a pendulum is released from rest at some height above the ground position.
It has only potential energy. B
In Figure B, a pendulum is still above the ground position, yet it is also moving.
It has BOTH potential energy and kinetic energy. C
In Figure C, a pendulum is at the ground position and moving with a maximum velocity.
It has only kinetic energy. D
In Figure D, the pendulum has reached the same height above the ground position as A.
It has only potential energy.

39. Energy consistently changes forms

40. Energy consistently changes forms
Am I above the ground?
Am I moving?
NO, h = 0, U = 0 J
Yes, v = 8 m/s, m = 60 kg
Position m v U K ME 1
60 kg
8 m/s
(= U+K)
0 J
1920 J
1920 J

41. Energy consistently changes forms
Energy Before
= Energy After KO
= U + K
= (60)(9.8)(1) + (.5)(60)v2
1920= v2
= 30v2
= v2
v = 6.66 m/s
Position m v U K ME 1
60 kg
8 m/s
0 J
1920 J 2
6.66 m/s
588 J
1332 J
1920 J

42. Energy consistently changes forms
Am I moving at the top?
No, v = 0 m/s
EB = EA
Using position 1
Ko = U
= mgh
=(60)(9.8)h
h = 3.27 m
Position m v U K ME 1
60 kg
8 m/s
0 J
1920 J 2
6.66 m/s
588 J
1332 J 3
0 m/s
1920 J
0 J

43. 6- Power
The Laws of Motion

44. Power Power is the time rate of energy transfer. The Power is given by
It is the amount of energy consumed per unit time
Power is the time rate of energy transfer.
The Power is given by
W F . d
P = =
t t
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second. So The unit of power is a Watt
where 1 watt = 1 joule / second

45. P = 33.33 watt Example 1 Solution: since, = = = W F . d 100 x 20 P
A 100 N force is applied to an object in order to lift it a distance of 20 m within 60 s. Find the power.
Solution:
since, = = =
P = watt
W F . d x 20 P
t t

46. Power Consumed: P = 2240 W Example 2 Solution: since, = = = =
What power is consumed in lifting a 70-kg robber 1.6 m in 0.50 s?
Solution:
since, = = = =
h =d and F = mg
W F . h m.g.h x10x 1.6 P
t t t
Power Consumed: P = 2240 W

47. Thank You for your Attention