•If two people mow two lawns of
equal size and one does the job
in half the time, who did more
work?
•If two people mow two lawns of
equal size and one does the job
in half the time, who did more
work?
•Same work. Different power
exerted.
POWER
POWER
the rate at which work is
done
the amount of work done per
unit time
w
N.m or J
f
d
N
m
w
P
J/s
fdcosθ (J)
t
s
Watt (W)
THE WATT
 A unit named after
Scottish inventor James
Watt.
 Invented the steam
engine.
 P = W/t
 Joules/second
 1 watt = 1 J/s
WATTS
 Used to measure power
of light bulbs and small
appliances
 An electric bill is
measured in kW/hrs.
1 kilowatt = 1000
W
HORSEPOWER (HP) =
745.5 WATTS
 Traditionally associated with engines.
(car,motorcycle,lawn-mower)
 The term horsepower was developed to quantify
power. A strong horse could move a 750 N
object one meter in one second.
750 N
Two physics students, Will and Ben, are in
the weightlifting room. Will lifts the 100pound barbell over his head 10 times in
one minute; Ben lifts the 100-pound
barbell over his head 10 times in 10
seconds. Which student does the most
work? Which student delivers the most
power?
WORD PROBLEM:
• A person weighing 600 N gets on an elevator.
The elevator lifts the person 6m in 10
seconds. How much power was used?
GIVEN
Force= 600 N
Distance= 6m
Time= 10s
FORMULA
P=w/t
SOLUTION and F.A.
WORD PROBLEM:
• A person weighing 600 N gets on an elevator.
The elevator lifts the person 6m in 10
seconds. How much power was used?
GIVEN
Force= 600 N
Distance= 6m
Time= 10s
FORMULA
P=w/t
SOLUTION and F.A.
P= (600N) (6m)
10s
=
3600 J
10s
=360 Watts (W)
WORD PROBLEM:
• How much time is needed to produce
720 Joules of work if 90 watts of power
is used?
GIVEN
w= 720J
P=90 W
t=?
FORMULA
SOLUTION and F.A.
WORD PROBLEM:
• How much time is needed to produce
720 Joules of work if 90 watts of power
is used?
GIVEN
w= 720J
P=90 W
t=?
FORMULA
w
t=
P
SOLUTION and F.A.
WORD PROBLEM:
• How much time is needed to produce
720 Joules of work if 90 watts of power
is used?
GIVEN
w= 720J
P=90 W
t=?
FORMULA
SOLUTION and F.A.
w
t=
720J
t=
P
90 W or 90J/s
t =8s
WORD PROBLEM:
• If 68 W of power is produced in 18
seconds, how much work is done?
GIVEN
t= 18s
P=68 W
w=?
FORMULA
SOLUTION and F.A.
WORD PROBLEM:
• If 68 W of power is produced in 18
seconds, how much work is done?
GIVEN
t= 18s
P=68 W
w=?
FORMULA
w= Pt
SOLUTION and F.A.
WORD PROBLEM:
• If 68 W of power is produced in 18
seconds, how much work is done?
GIVEN
t= 18s
P=68 W
w=?
FORMULA
w= Pt
SOLUTION and F.A.
w= (68 J/s) (18s)
w=1224 J
WORD PROBLEM:
• A tired squirrel (mass of approximately 1 kg) does push-ups by
applying a force to elevate its center-of-mass by 5 cm in order to
do a mere 0.50 Joule of work. If the tired squirrel does all this
work in 2 seconds, then determine its power.
GIVEN
FORMULA
SOLUTION and F.A.
WORD PROBLEM:
• A tired squirrel (mass of approximately 1 kg) does push-ups by
applying a force to elevate its center-of-mass by 5 cm in order to
do a mere 0.50 Joule of work. If the tired squirrel does all this
work in 2 seconds, then determine its power.
GIVEN
W= 0.50 J
T= 2.0seconds
FORMULA
SOLUTION and F.A.
WORD PROBLEM:
• A tired squirrel (mass of approximately 1 kg) does push-ups by
applying a force to elevate its center-of-mass by 5 cm in order to
do a mere 0.50 Joule of work. If the tired squirrel does all this
work in 2 seconds, then determine its power.
GIVEN
W= 0.50 J
T= 2.0seconds
FORMULA
P =W / t
SOLUTION and F.A.
WORD PROBLEM:
• A tired squirrel (mass of approximately 1 kg) does push-ups by
applying a force to elevate its center-of-mass by 5 cm in order to
do a mere 0.50 Joule of work. If the tired squirrel does all this
work in 2 seconds, then determine its power.
GIVEN
W= 0.50 J
T= 2.0seconds
FORMULA
P =W / t
SOLUTION and F.A.
P = (0.50 J) / (2.0 s)
P= 0.25 Watts
WORD PROBLEM:
• A student uses 150N to push a block of
wood up a ramp at a constant velocity of
2.6 m/s.What is his power output?
GIVEN
F = 150 N
V = 2.6 m/s
FORMULA
SOLUTION and F.A.
WORD PROBLEM:
• A student uses 150N to push a block of
wood up a ramp at a constant velocity of
2.6 m/s.What is his power output?
GIVEN
F = 150 N
V = 2.6 m/s
FORMULA
P= F * v
SOLUTION and F.A.
WORD PROBLEM:
• A student uses 150N to push a block of
wood up a ramp at a constant velocity of
2.6 m/s.What is his power output?
GIVEN
F = 150 N
V = 2.6 m/s
FORMULA
SOLUTION and F.A.
P= F * v P= (150N) (2.6
m/s)
P= 390Watts