Younes Sina's Lecture at Pellissippi State Community College

1. Chapter 6
Work and Energy
Younes Sina

2. Work (W): Work is defined as the product of parallel force and
distance. W = F ll x .
F ll denotes the component of force F that is parallel to
displacement , x .
The SI unit for work is N.m called " Joule", and the American unit is
" lb-ft "
If the direction of F is parallel to that of x, the work done is simply
W = F x.

3. Example :
Find the work done by force F = 25N in
pushing the block a distance of 14m .
25 N force applied
Block moved by 14 m

4. 25 N
14 m
W=F . x
W=(25N)(14 m)
W=350 Joules
Solution

5. If an object placed on a horizontal surface is pushed or
pulled by force F as shown below, the component of F that
is parallel to x does useful work .
The component perpendicular to x does no work.

7. Example : A block is pulled a distance of x = 24m
from A to B as shown via force F = 45N that makes
a 30 .0 o angle with the horizontal surface . Calculate
the work done by F.
x = 24m
30 .0 o

8. Solution: For θ = 30.0o, and F = 45N
F ll = 45N cos (30 .0 ) = 39N
W = F ll x
W = (39N)(24m) = 940 Nm
W = 940 J

9. Example : In the following figure, find
(a) the magnitude of F such that the block slides at constant
speed to the right .
(b) Find the work done by this force if the displacement is 38 m .

11. Example: In previous example, find the work done by
the frictional force, F k , within the same distance.

12. Solution: F k acts opposite to x; therefore, it does negative
work.
W friction = F k x
W = ( -45N )( 38m ) = -1700 J

13. Energy: Energy is defined as the ability to do work. Energy and
work are expressed in same units. Typical units are: J, cal, kcal, and eV.
Types of Energy:
Energy exists in different forms such as mechanical, electric, nuclear,
light, chemical, etc... .
In this chapter, mechanical energy will be discussed only.
Heat energy is also a will be discussed in Chapter 12 .
Mechanical Energy:
Kinetic energy, Gravitational potential energy, and elastic (spring)
potential energy are forms of mechanical energy and will be
discussed under this topic.

14. I ) Kinetic Energy (K.E.):
Kinetic energy is the energy that a mass (an object) has because of
its motion. As long as an object moves or has some speed, it has
kinetic energy.
K.E. is proportional to mass ( M) and proportional to the square of
velocity ( v 2 ) .
K.E. = (1/2) Mv 2
II ) Gravitational Potential Energy (P.E.):
This energy is the type that an object has due to its elevation with
respect to a reference level. This energy is significant when objects
are in the vicinity of planets, stars and other heavenly masses .
Gravitational P.E. is proportional to the acceleration of gravity ( g) of
the planet or star, the mass of the object, M, and its elevation ( h)
from a reference level.
P.E. = Mgh

15. Example :
Calculate the K .E . of a 2000-kg car that is moving
(a) at 10.0m/s
(b) 20.0 m/s
(c) 30.0 m/s

16. Solution: (3 sig. fig.)
(a) (K.E.)1 = (1/2)(2000.kg)( 10 .0m/s) 2 = 100,000 (kg m/s 2)m = 100,000 J
(b) (K.E.)2 = (1/2)(2000.kg)( 20 .0m/s) 2 = 400,000 (kg m/s 2)m = 400,000 J
(c) (K.E.)3 = (1/2)(2000.kg)( 30 .0m/s) 2 = 900,000 (kg m/s 2)m = 900,000 J

17. Example:
A rock of mass 6.1 kg is raised to the top of a building 13 m high.
Calculate its gravitational P .E . with respect to the ground.

18. Solution:
P.E. = Mgh
P.E. = (6 .1 kg)(9 .8 m/s 2)(13 m)
P.E. = 780 Nm = 780J

19. Example :
In the following figure, a 95-kg crate is pushed up on an incline that
is practically frictionless. The incline is 8 .0m long and makes a 31 o
angle with the horizontal floor. Calculate the P .E . Of the crate with
respect to the floor when it reaches the top of the incline.

21. Example : A swimming pool is on the top of a hill at an average elevation of 150m
from a certain ground level . The swimming pool has dimensions:
15 m X 25 m X 2 .4 m and is full. The mass density of water is 1.000 ton/m 3.
Determine
(a) the P .E . Of the water in the pool with respect to that level .
If the pool is allowed to empty to the ground level,
(b) how much energy is at most available for use?
(c) how much electric energy will become available? If a generator uses this energy
for electricity production, and the overall efficiency is 62%

22. Solution:
(a) The pool's volume is V = w l h
V = 900m 3
r = M / V
M = r V
M = (1000 kg/m 3)(900 m3) = 900,000 kg
P.E. = Mgh
P.E. = ( 900,000 kg )( 9 .8 m/s 2 )( 150m ) = 1.3 × 10 9 J
(b) 1.3 × 10 9 J
(c) 0.62 (1.3 × 10 9 J) = 8.2 × 10 8 J
100 W
2278 hr

23. Example : A 750-kg car is traveling at a velocity of 72 km/h eastward
and on a level road . Determine
(a) its initial kinetic energy (K .E .) i
If this car is slowed down to a speed of 36 km/h, calculate
(b) its final (K.E.) f .
(c) How much is the change in its kinetic energy and where does this
energy go ?
Vi=72 km/h Vf=36 km/h

24. Solution:
(a) (K.E.) i = (1/2)Mv i
2
(K.E.) i = (1/2)( 750 kg)( 20 . m/s)2 = 150,000 J
(b) (K.E.) f = (1/2)Mv f
2
(K.E.) f = (1/2)( 750 kg)( 10 . m/s)2 = 38,000 J
(c) Δ(K.E.) = (K.E.) f - (K.E.) i
Δ(K.E.) = 38,000 J - 150,000 J = - 112,000 J
This energy is consumed by force of friction (brakes force, for
example) that acts opposite to the direction of motion . In fact, it is
the work done by force of friction. Most of this energy converts to
heat and warms up the brakes. Friction does negative work .

25. Example: If the work done by force of friction is -112,000J .
Use the work formula to calculate the force applied by friction if brakes were used
within a distance of 56m .
Wf=-112,000J
56m

26. Solution:
W = F k x
- 112,000 J = F k (56m)
F k = - 2000N

27. Work-Kinetic Energy Theorem:
The work done by the net force acting on a mass is equal to the
change in the kinetic energy of that mass.
(ΣF ) x = Δ(K.E.)
(ΣF ) x = (K.E.) f - (K.E.) i

28. Example : A 900-kg car traveling at 15m/s changes its speed to
25m/s in a distance of 50 m due to a net force . Calculate
(a) the net force
(b) the engine force if the frictional forces add up to 1400N .
V=15m/s V=25m/s
X=50 m
M= 900 kg

29. Solution:
(a) Work-K.E. theorem :
(Σ F) ∙ x = (K.E.) f - (K.E.) i
Σ F (50 m) = (1/2)(900)(25) 2 - (1/2)(900)(15) 2
Σ F = 3600 N
(b) Σ F = F e - Ff
3600N = Fe - 1400N
F e = 5000 N

30. Example : A boy pulls a 45.0-kg sled (including his friend in it)
from rest for a distance of 12.0 m with a horizontal and constant
force of 165.0 N. The frictional force between the sled and snow
is 55.0 N. Calculate
(a) the work done by the boy on the sled
(b) the work done by the frictional force on the sled
(c) the work done by the net force on the sled
(d) the speed of the sled at the end of the 12.0 m distance .
12.0m
45.0 kg
165.0 N
Wf=55.0 N
v=0 m/s v=? m/s

31. Solution:
(a) Wboy = F ll x
W = (165 N)(12 .0 m) = 1980 J
(b) Wfriction = F k x
W = (55 .0 N)(12.0 m) = 660.0 J
(Work of friction is negative)
(c) Wnet = Fnet x
(Σ F) x = (165N - 55.0 N)(12.0 m) = 1320 J
(d) (Σ F) x = (K.E.) f - (K.E.) i
1320 J = (1/2)(45 .0kg) Vf
2 - (1/2)(45 .0kg) (0)2
Vf = + 7.66 m/s

32. The Law of Conservation of Energy
This law states that:
"Energy is conserved.
It is neither created nor destroyed.
It converts from one form to another".

33. Conservation of Mechanical Energy
Example : In the following figure, neglecting friction, find the speed
Of the 750-kg car at the bottom of the hill . Suppose the car is put in
neutral and starts from rest from the top of the hill .
V=0, M=750 kg
45 m

34. Conservation of mechanical energy:
Total Energy at A = Total Energy at B
(K .E .)A + (P .E .)A = (K .E .)B + (P .E .)B
(1/2)MVA
2 + M g hA = ( 1/2)MVB
2 + M g hB
VA = 0 and hB = 0
M g hA = ( 1/2)MVB
2
g hA = (1/2)VB
2
2 g hA = VB
2
VB = 30 m/s
A
BhA
hB = 0
VB

35. Example :
In the following figure, if 120,000 J of energy is consumed by
frictional forces, find the speed of the 750-kg car at the bottom
of the hill . Suppose the car is put in neutral and starts from rest
from the top of the hill .
frictional Energy=120,000 J
In this case, part of the available energy will be wasted by friction

36. Conservation of mechanical energy:
Total Energy at A - W friction= Total Energy at B
(K .E .)A + (P .E .)A - Wfriction = (K .E .)B + (P .E .)B
(1/2)M A
2 + M g hA -120,000J = ( 1/2)MV B
2 + M g hB
V A = 0 and h B = 0
M ghA -120,000J = ( 1/2)MV B
2
(750)(9 .8)(45) J - 120,000 J = 0 .5(750 kg)VB
2
562(m/s) 2 = VB
2
VB = 24 m/s
V=0, M=750 kg

37. Example: In the figure shown, find
(a) the work done by the frictional force on the car as it coasts
down the hill in neutral
(b) the energy loss due to friction
(c) the speed of the 750-kg car (VB) at the bottom of the hill .
400 m

38. Solution:
(a) The work done by friction is
W friction = F k x
W friction = ( -300N)(400m) = -120,000 J
(b) -120,000 J
(c) Conservation of mechanical energy:
Total Energy at A - Wfriction = Total Energy at B
(K .E .)A + (P .E .)A - Wfriction= (K .E .)B + (P .E .)B
(1/2)MVA
2 + M g h A - 120,000 J = (1/2)MVB
2 + M g hB
VA = 0 and hB = 0
M ghA -120,000 J = (1/2)MVB
2
(750)(9.8)(45) J - 120,000 J = 0.5(750)VB
2
[562(m/s)]2 = VB
2
VB = 24 m/s

39. Power is defined as the work done per units of time.
Mathematically:
The (SI) unit for power is :
The industrial unit: hp
Power
t
W
P


=
Wattcalled
s
J
s
ftlb
hphP

== 5501andwatts7461

40. Example : An electric motor is capable of delivering 7 .2 Million
Joules of work in one hour and 400.0 seconds. Find the power
of the motor in watts, kilowatts, and hp .
Solution:
P = W/t
P = (7 .2 x10 6 J )/(4000 s) = 1800 J/s = 1800 watts
P = 1.8 kilowatts (kw)
P = (1800/746 ) hp
P = 2.4 hp

41. Example:
Calculate the amount of work or energy that a 4.50 hp
electric motor can deliver in 10.0 hours.
Solution:
P = 4 .5hp = (4 .50)(746 watts) = 3360 watts or ( J /s )
P = W / t
W = Pt
W = (3360 J/s)(36,000s) = 1.21 x 10 8 J

42. Efficiency: When a device receives power from a source, it does not deliver all of
it in the intended form and converts a portion of it to other undesired forms. This
makes a device to be less than 100 percent efficient.
Most power plants have efficiencies of under 50 percent. A good nuclear power
plant is about 45% efficient . Early coal plants were hardly 20% efficient .
Eff. = Pout /Pin

43. Example : A 1.25 hp electric pump with an efficiency of 92.5% is used to perform
1 .40 Mega Joules of work.
(a) How long does it take for the electric pump to do the job ?
(b) If it was 100% efficient , how long would it take?
Solution:
A 1.25 hp electric pump means it draws 1 .25 hp of electric power from the
source.
It generally cannot give back all of it .
Make sure that you write the formulas with horizontal fraction bars .
Eff. = P out / P in
0 .925 = P out / (1 .25 hp)
P out = 0 .925(1 .25hp) = 1.16 hp
(a) P = W / t
t = W / P
t = (1 .40 x 10 6 J) / ( 1 .16 x 746 watts ) = 1620 sec
(b) P = W/t
t = W/P
t =(1 .40 x 10 6 J)/( 1 .25 x 746 watts ) = 1500 sec

44. Example: Calculate
(a) the necessary power to empty a swimming pool with dimensions
(12m)(25m)(2 .4m) by pumping its water to an average height of 3.5m in 2.5hrs .
The mass density of water is 1000 kg/m 3.
(Hint: The power you calculate is the power that the electric pump has to deliver
to water, Pout) .
If the efficiency of the electric pump used is 88% , calculate
(b) the electric power that the electric motor pulls from the power source, Pin .

45. Solution:
V = w. l. h
V = 720m3
r = M / V
M = r V
M = (1000 kg/m 3)(720 m3) = 720,000 kg
(a) P = W / t
W = F x
F = the weight of water= Mg
x = height
P = (F x) / t
P = ( Mg x) / t
P = (720,000kg )( 9 .8 m/s 2 )( 3 .5m ) / (2 .5 X 3600s ) = 2700 watts
P = 2.7kw
P = (2700 / 746) hp = 3.6 hp
(b) Eff. = P out / P in
0 .88 = 2700watts / P in
P in = 2700watts / 0 .88 = 3100 watts

46. Another Version of Power Formula:
P = W / t
W = F x
P = (F x) / t
(x / t )= v
P = F v
This means that power is equal to force times velocity (if both of
force and velocity are in the same direction).
As you see the product of velocity and force is a constant if a
constant power is available . When P is constant, the product F ∙v
is constant . When a greater F is needed, a smaller v is attainable
and vice versa .

47. Example :
A car is using 56 hp of its total power. When it is going up a steep hill, a force of
4200 N is needed for constant velocity motion of it up that hill. When it is
traveling along a level and horizontal road, a force of 1200 N is needed for its
constant velocity motion. Find the velocity of the car in each case .
P = F v
56 hp
56 hp
F= 4200 N
F=1200 N

48. Solution:
P = (56)(746 watts) = 42000 watts
(a) P = F v
v = P / F
v = 42000 watts / 4200N = 10 m/s
(b) P = F v
v = P / F
v = 42000watts / 1200N = 35 m/s

49. problems 1 through 5 chapter 6