Work, Power, & Efficiency
November 2015
Work
 Work: The word looks the same, but it has different
meaning in physics than the way it is normally used
in the everyday language
Work W is done when a constant force F
exerted on an object through distance d
F
θ
F
Fd
θ
Fd
d
 Only the component of force that acts in the same direction as
the motion is doing work on the box.
 Vertical component is just trying (unsuccessfully) to lift the object up.
work = force along distance × the distance moved
W = Fdd = Fd cosθ
work = force × distance moved × cos of the angle between them
Units
 The SI unit for work is the newton–metre and is called
the joule named after the 19th Century physicist
James Prescott Joule.
 1 J (Joule) = 1N x 1 m
 Work is a scalar (add like ordinary numbers)
A force is applied. Question: Is the work done by that force?
Work - like studying very hard, trying to lift up the car and get
completely exausted, holding weights above head for half an hour is
no work worth mentioning in physics.
 According to the physics definition, you
are NOT doing work if you are just
holding the weight above your head
(no distance moved)
 you are doing work only while you are
lifting the weight above your head
(force in the direction of distance moved)
Who’s doing the work around here?
NO WORK
WORK
 If I carry a box across the room I do not do work on it because
the force is not in the direction of the motion (cos 900 = 0)
θ = 00
00< θ <900
θ = 900
900< θ <1800
cos θ = 1
cos θ = +
cos θ = 0
cos θ = –
θ = 1800
cos θ = –1
Work done by a force F is zero if:
 force is exerted but no motion is involved:
no distance moved, no work
 force is perpendicular to the direction of motion (cos 900 = 0)
d
d
F
F
F
motion
normal
force
tension in
the string
gravitational
force
Work done by force F is:
 positive
when the force and direction of motion are generally in the same
directions
00< θ < 900 → cos θ = +
cos 00 = 1
W = Fd
 negative
when the force and direction of motion are generally in the
opposite directions
900< θ < 1800 → cos θ = –
cos 1800 = –1
W = - Fd
(the work done by friction
force is always negative)
We do:
1) Is work positive, negative, or zero?
a) a book falls off a table
b) a rocket accelerates through space
2) Mike is cutting the grass using a human-powered lawn
mower. He pushes the mower with a force of 45 N directed at
an angle of 41° below the horizontal direction. Calculate the
work that Mike does on the mower in pushing it 9.1 m across
the yard
We do:
1) Is work positive, negative, or zero?
a) a book falls off a table
Postive! The force (gravity) is in the same direction
as the motion.
b) a rocket accelerates through space
Positive! The force (of expelled gasses on the
rocket) are in the same direction as the motion.
We do:
2) Mike is cutting the grass using a human-powered lawn
mower. He pushes the mower with a force of 45 N directed at
an angle of 41° below the horizontal direction. Calculate the
work that Mike does on the mower in pushing it 9.1 m across
F = 45 N
the yard
d
410
F
W = Fd cos θ = 310 J
d = 9.1 m
θ = 410
You do:
1) Is work positive, negative, or zero?
a) A waiter carries a tray full of meals above his head by
one arm straight across the room at constant speed.
b) friction slows a sliding hockey puck
2) Forward force is 200 N. Friction force is 200 N. The
distance moved is 200 km. Find
a) the work done by forward force F on the car.
b) the work done by friction force Ffr on the car.
c) the net work done on the car.
You do:
1) Is work positive, negative, or zero?
a) A waiter carries a tray full of meals above his head by
one arm straight across the room at constant speed.
No work! The force is exerted upward, and the motion
is horizontal.
b) friction slows a sliding hockey puck
Negative work! The force is opposite the motion.
Forward force is 200 N. Friction force is 200 N.
The distance moved is 200 km. Find
a. the work done by forward force F on the car.
b. the work done by friction force Ffr on the car.
c. the net work done on the car.
F = 200 N
a. WF = Fd cos 00
Ffr = 200 N
b. Wfr = Ffr d cos 1800 = - 4x107 J
d = 2x105 m
c. the net work done on the car means the
work done by net force on the car.
It can be found as:
= 4x107 J
W = WF + Wfr = 0
or
W = Fnet d cos θ = 0
(Fnet = 0)
Work done by a varying force - graphically
 W = Fd cos θ applies only when the force is constant.
 Force can vary in magnitude or direction during the
action.
 Examples: 1) rocket moving away from the Earth – force
of gravity decreases 2.) varying force of the golf club on
a golf ball, etc …
 In these cases, work done is most easily determined
graphically.


The lady from the first slide is pulling the car for 2 m with force of 160 N
at the angle of 60o , then she gets tired and lowers her arms behind her at
an angle of 45o pulling it now with 170 N for next 2 m. Finally seeing the
end of the journey she pulls it horizontally with the force of 40 N for 1 m.
Work done by her on the car is:
W = (160 N)(cos 60o)(2m) +(170 N)(cos 45o)(2m) + (40 N)(cos 0o)(1m)
W = 80x2 + 120x2 + 40x1 = pink area + green area + blue area = 440 J
http://www.kcvs.ca/map/java/applets/workE
nergy/applethelp/lesson/lesson.html#1
In general:
The area under a Force - distance graph equals
the work done by that force
Graphical determination of work: You do
A man pushes a shopping cart 6 meters. What is the work done?
Find area: (1/2)(2)(1)+(2)(2)+(1/2)(2)(3) = 8 J
Power
◘ Power is the work done in unit time or energy converted in
unit time
P=
W
t
or P =
E
t
measures how fast work is done or how quickly energy is
converted.
Units:
1 W(Watt) =
1J(joule)
1s
A 100 W light bulb converts electrical energy to heat
and light at the rate of 100 J every second.
Calculate the power of a worker in a supermarket who
stacks shelves 1.5 m high with cartons of orange juice,
each of mass 6.0 kg, at the rate of 30 cartons per minute.
Calculate the power of a worker in a supermarket who
stacks shelves 1.5 m high with cartons of orange juice,
each of mass 6.0 kg, at the rate of 30 cartons per minute.
W
Fdcos00 (30×60N)×1.5m
P=
=
=
t
t
60s
P = 45 W
Efficiency
◘ Efficiency is the ratio of how much work, energy or power
we get out of a system compared to how much is put in.
useful output
efficiency =
total input
eff =
Wout
E
P
= out = out
Win
Ein
Pin
◘
No units
◘
Efficiency can be expressed as percentage by multiplying
by 100%.
No real machine can ever be 100% efficient,
because there will always be some energy
lost as heat
◘
A car engine has an efficiency of 20 % and produces an
average of 25 kJ of useful work per second.
How much energy is converted into heat per second.
A car engine has an efficiency of 20 % and produces an
average of 25 kJ of useful work per second.
How much energy is converted into heat per second.
Eout
eff =
Ein
0.2 =
25000J
Ein
Ein = 125000 J
heat = 125 kJ – 25 kJ = 100 kJ
Quick Review (2 min)
With your table partner …
Person with longer hair:
• Define power in your own words
• State the equation and units for power.
Person with shorter hair:
• Define efficiency in your own words
• State the equation and units for efficiency
Power & Efficiency – You do
 An elevator lifts a total mass of 1.1X103 kg a
distance of 40.0 m in 12.5 s. How much power
does the elevator deliver?
P = W/t = Fd/t = mgd/t = 1100*9.8(40.0)/12.5 = 34,500 W
 What work is required to lift a 215 kg mass a
distance of 5.65 m, using a machine that is
72.5% efficient?
E = Wout / Win => Win = Wout / E
Win = 215*5.65*9.8 / .725 = 16, 400 J