# Power_

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```					          Power!

“Work” Refresher
• Remember that the formula for finding the
amount of work done upon an object is:
W = (F)(d)(cos Θ)
•   W = the work done
•   F = the force required to cause displacement
•   d = displacement of the object
•   Θ = the angle between the direction of the force
and the direction of the displacement
Definition of Power
•Common definition usually includes
•Physics definition - the timed rate at
which work is done
•depends directly upon the work and
inversely upon the time to do that work
If you increase work, then you increase power
BUT
If you increase time, then you decrease power
How “Hard” Are You Working?

• The rate at which work is done is called
power:

(Power) = (Work Done) / (Time Spent Working)

P=W/t

• Power is “how hard”
something is working.
Units for power:

1J
=1W   one watt (W) equals one
1s
joule (J) per second (s)

joule per second = watt
Kilowatts and Horsepower
• Another common unit of power (not used in
science, but used in everyday life) is the
“horsepower” – basically the rate at which a
(very powerful, very healthy) horse can do work
over a 10 hour “work day.”
• 1 horsepower (hp) = 746 W or 0.746 kW
• We say 1,000 watts (W) = 1 kilowatt (kW), the
same way that 1,000 meters = 1 kilometer
How fast can your car do work?
• A compact car may have a 120 hp
engine.
• That means the car’s power is (120
hp)(746 W/hp) = 89,500 W
• So a typical compact car
can do 89,520 J worth of
work each second.
Remember this Example of Work?
• You are pushing a very heavy stone block
(200 kg) across the floor. You are exerting
620 N of force on the stone, and push it a
total distance of 20 m before you get tired
and stop.
How much work did you just do?

W = (620 N)(20 m)(cos 0°) = 12,400 J
Power Example
Let’s say that it took us 40 s to move that 200 kg
stone block the 20 m.
– Remember that we did 12,400 J of work on the stone
block
– Given Information:
• W = 12,400 J
• t = 40 s
– Basic Equation:
• P = W/t
– Working Equation:
• P = 12,400 J / 40 s
• P = 310 J/s
– Since it took us 40 s to move the block, we were
doing 310 J of work per second OR generating 310 W
of power
Remember the Weightlifter?
• A weightlifter lifts a barbell with a mass of
280 kg a total of 2 meters off the floor.
What is the minimum amount of work the
weightlifter did?
– The barbell is “pulled” down by gravity
with a force of (280 kg)(9.8 m/s2) =
2,744 N
– So the weightlifter must exert at least
2,744 N of force to lift the barbell at all.
– If that minimum force is used, the
work done will be:
W = (2,744 N)(2 m)(cos 0°) = 5,488 J
Power
– How much power does the weightlifter
generate if it takes him 2 seconds to lift the
weight? (W = 5,488 J)
– P = 5,488 J / 2 s
– P = 2,744 J/s OR 2,744 W
– We say 1,000 watts = 1 kilowatt
– So the weightlifter’s power was 2.7 kW
Fast work isn’t more work
• Go back to our 200 kg block example.
Remember that when it took us 40 s to
push the block the 20 m when we applied
a force of 620 N, that implied that we had
a power output of 310 W.
• If we exerted the same force (620 N) and
pushed the block the same distance (20
m), but took half as long to do so (20 s),
our power output would double to 620 W.
Fast work isn’t more work
• *But* notice that the total work done
doesn’t change – we still exerted 620 N of
force over a distance of 20 m.
• So increasing power output doesn’t
mean you’re doing more work, it means
you’re doing the work faster.
Power Humor
• Who is the most powerful teacher on this
planet?
– Ms. Watts

• What did the baby light bulb say to the
momma light bulb?
– I wuv you watts!
Game Time!!!
• Break into 4 groups (with even amounts of
people in each group)
• A question will be projected onto the board
and 2 groups will race to complete the
question correctly
• The winning group will get a reward
• In order to receive the reward, groups
must include all steps needed to solve a
word problem
then Groups 2 and 4 will work question 2,
etc
• Groups not working on questions, must
stay quiet
• Rewards will go to those actively
participating with their group (writing on
the board, working it out on separate
paper, paying attention, etc)
Question 1 – Groups 1 and 3
• A box that weighs 575 N is lifted a
distance of 20.0 m straight up. The job is
done in 10.0 s. How much power is
generated?
• Include a diagram, given information, B.E.,

• Answer: 1150 W or 1.15 kW
Question 2 – Groups 2 and 4
• An electric motor develops 65 kW of
power as it lifts a loaded elevator 17.5 m in
35 s. How much force (FL) does the motor
exert?
• Include a diagram, given information, B.E.,

Question 3 – Groups 1 and 4
• Jose lifts a 20 kg mass to a height of 2.0 m
in 5.0 s. Sue lifts 30 kg to a height of 1.5
m in 8.0 s. Who did more work? Who had
more power?
• Include a diagram, given information, B.E.,
– Sue did more work (450 J), less power (56 W)
– Jose generated more power (80 W), less work
(400 J)
Question 4 – Groups 2 and 3
• A refrigerator is loaded into a moving van
by a pushing it up a 10.0 m ramp at an
angle of inclination of 11.0° in 20 s with a
force of 850 N. How much work is done
by the pusher? How much power was
generated?
• Include a diagram, given information, B.E.,
• W = 8500 J
• P = 425 W
And the WINNER is…
• Everyone!!
• Because everyone helped to work the
problems you have all been preparing for
the test
• Good job!

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