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Work, Energy and Power
Work
W = Fs
• W= work
• F = force
• s = displacement
• Whe a force makes an object move, work has
been done.
• Whenever work is done, energy is transferred
form one place to another or transformed from
one form to another.
• Units are = 1Nm = 1J(joule)
Practice Problem 1
• A 150N force moves an object through a
displacement of 6m. Calculate the work
done.
• Answer: W=Fs=150 x 6 = 900J
• Work isn’t always done in a stratight line.
That means we have to take into account the
direction of the displacement. This is done
by adding “cosθ” to the equation.
• W = F s cosθ
• This means that work done is only in the
same direction of the force vector.
Practice Problem 2
• A force F = 100N is used to move a box
though a horizontal distance of 14m. If the
force is applied to the box at angle θ = 35°
as shown, calculate the work done on the
box.
• Answer: 1147J
• This equation is only applicable if the force is
constant.
• In real life there are lots of non-constant forces.
• We can use a graph to determine work done.
• On a Force/Displacement graph the area under
the graph is the work done.
ENERGY!!!!
There are lots of different forms of energy.
1. Kinetic
2. Gravitational potential
3. Elastic potential
4. Electrical potential
5. Light(radiation)
6. Nuclear(mass)
7. Chemical potential
8. Sound
9. Thermal
Kinetic
• Ek of an object is numerically equal to the work
done.
• Associated with objects in motion
• Depends on both an objects
speed and mass
• It is a scalar quantity
Kinetic
•
•
•
•
Kinetic energy = 1/2 *mass*speed2
Ek = W = 1/2 mv2
The unit of KE is joule(J)
If the speed is doubled the energy is
quadrupled.
• Ek can also be defined in terms of the
momentum of an object.
• Ek = p2 / 2m
Kinetic
• A bowling ball and a volleyball roll at the
same speed.
• Which has more kinetic energy?
Kinetic
• A 7kg bowling ball moves at 3 m/s.
How fast must a 2.45 g ping pong ball
move to have the same kinetic energy
as the bowling ball?
• KE=1/2 mv2
Practice Problems:
• Determine the kinetic energy of a 625-kg roller coaster car
that is moving with a speed of 18.3 m/s.
• If the roller coaster car in the above problem were moving
with twice the speed, then what would be its new kinetic
energy?
• Missy Diwater, the former platform diver for the Ringling
Brother's Circus, had a kinetic energy of 12 000 J just prior
to hitting the bucket of water. If Missy's mass is 40 kg,
then what is her speed?
• A 900-kg compact car moving at 60 mi/hr has
approximately 320 000 Joules of kinetic energy. Estimate
its new kinetic energy if it is moving at 30 mi/hr. (HINT: use
the kinetic energy equation as a "guide to thinking.")
Answers:
A.105,000J
B.419,000J
C.24.5m/s
D.80,000J
Practice Problem 3
• Calculate the kinetic energy of a 110kg
rugby player running the 100m in time 15s.
• Answer: Vavg = Δs/Δt = 100/15 = 6.67m/s
• Ek = ½ mv2 = 0.5 x 110 x 6.672
• Ek = 2440J
Practice Problem 4
• Calculate the momentum of the rugby
player in the previous example using Ek =
p2 / 2m
• Answer:
• Ek = p2 / 2m
Potential
• Associated with an object that has the
potential to move because of its position
• Depends on the interaction with its
environment
• store energy as the
result of its position
Potential
• Gravitational potential energy
– Potential energy due to gravity
• the energy stored in an object as the
result of its vertical position or height.
• ΔEp = mass*free-fall acceleration*height
• ΔEp = mgΔh
Potential
• ΔEp = mgΔh
• Notice that ΔEp is dependant on free-fall
acceleration being constant.
• Also notice that g and h aren’t properties of
the actual object.
Potential
• ΔEp = mgΔh
• The higher that an object is elevated, the
greater the gravitational potential energy
• a doubling of the height will result in a
doubling of the gravitational potential energy
Potential
• Is it possible to have a negative
potential energy?
• Can an object have both a positive
potential energy and a negative
potential energy?
Potential
• How is h defined?
– Relative to a “zero”
• What can be a “zero”?
– Anything can be defined as “zero”
• Discussion: A ball that falls from one
building rooftop to another buildings
rooftop. Where is zero?
Practice Problem 5
• Calculate the gain in potential energy of a
70kg person climbing 5 flights of stairs
through a vertical height of 20m.
• Answer:
• ΔEp = mgΔh = 70 x 10 x 20 = 14000J
Principle of Conservation of Energy
• Energy cannot be created or destroyed. I
simply changes from one form to another.
• Ebefore = Eafter
• Ek = Ep
• ½ mv2 = mgh
Practice Problem 6
• An object falls from a height of 25m. Calculate
its velocity as it hits the ground.
• Use energy conservation, not the 4K equations.
• Ebefore = Eafter
• Ek = Ep
• ½ mv2 = mgh
•
• Answer:
• v2 = 2gh = (2 x 10 x 25)
• v = 22.4m/s
Connecting Momentum and Energy
Practice Problem 7
At a railway station, two trains slowly collide
in order to “couple” (join together). If one
train of mass 12000kg is moving at 0.75m/s
and the other of mass 3000kg is stationary,
calculate the velocity of the two as they
move away.
We already solved this for .6m/s. Now tell me
the kinetic energies before and after the
collision.
Practice Problem 9
• A boy of mass 30 kg is being given a lift on the back of a 10kg
bicycle by a girl of mass 40kg. They are travelling at a steady speed of
2.5m/s.
• The boy wishes to get off the back of the bicycle while it is still
moving.
• (PART A)He knows that if he just puts his feet on the ground and
stands up he is likely to fall over. Explain why.
• So instead he pushes himself off the back of the bicycle by pushing
forward on the bicycle frame with his hands, so that he lands on the
ground with zero horizontal velocity.
• (PART B) Calculate the velocity of the bicycle and the girl
immediately after the boy has left the bicycle.
• (PART C) Calculate the total kinetic energy of the system (bicycle
and both children) before and after the boy gets off. Explain the
reason for any difference.
Answer:
PART A: His feet will stop when they hit the
ground but his body will still be travelling
forward at the bike speed, so he will fall
over forwards.
PART B: 4m/s
PART C: before – 250J
after – 400J
The increase in KE is due to energy provided
or work done by the boy pushing on the
bicycle.
POWER!!!
POWER!!!
• is the rate at which work is done
• or the rate at which energy is consumed
• P = W/t
So….
• The more power you have the more
work you can do in the same time.
• The more power you have you can do
the same amount of work in a shorter
time.
• There is an inverse relationship
between work and power
How did we get there?
Remember: P = W/t
Quick review: what is work?
• W = Fd
Quick review: what is velocity?
• v = Δx/Δt or v = d/t
• So P = F (d/t)
• Or P = F v
POWER!!!
• P=Fv
• This equation shows us that a powerful
machine is both strong (big force) and
fast (big velocity).
– Ex. A powerful car engine is strong and
fast.
That was a lot to take in
• Equations for power are:
– P = W/t
– P = F (d/t)
–P=Fv
– P = ΔE/t
What units??
The SI unit of power is the watt.
• 1 watt is = 1 joule / second
• W=J/s
What units??
Horsepower is also a unit.
• 1 horsepower = 746 watts
• Hp = 746W
Examples of power
•
•
•
•
A dim light bulb 40 W
A really bright bulb 500 W
Indoor Christmas light .7 W
Outdoor Christmas light 7 W
Examples situation
• A 193kg curtain needs to be raised
7.5m, in 5s. You have 3 motors with
power ratings 1.0kW, 3.5kW and 5.5kW.
Which motor is best for the job?
• How much time would it take for
each motor to do the same amount of
work?
Examples situation
• Two horses pull a cart. Each exerts a
force of 250.0 N at a speed of 2.0 m/s
for 10.0 min.
• 1) What is the power delivered by the
horses?
• 2) How much work is done by the two
horses?
Practice:
• Two physics students, Will N. Andable and Ben Pumpiniron,
are in the weightlifting room. Will lifts the 100-pound 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?
• During a physics lab, Jack and Jill ran up a hill. Jack is twice as
massive as Jill; yet Jill ascends the same distance in half the
time. Who did the most work? Who delivered the most power?
(plug in fake numbers)
• When doing a chin-up, a physics student lifts her 42.0-kg body
a distance of 0.25 meters in 2 seconds. What is the power
delivered by the student's biceps?
• Mr. B gets bored after school one day and decides to play in
the hall. He sits in his rolling chair and pushes off the wall with
12N of force producing 30W of power. What was his resulting
speed he traveled down the hallway?
Practice Problem 9
• Calculate the power generated in the legs of
a 60kg student climbing a flight of stairs
through a vertical height of 8m in a time of:
• 1.2s
b) 0.75s
• *Hint: Think energy change.
• Answer:
• A) Power = W/t = mgh/t = (60*10*8)/1.2 = 4000W
B) Power = W/t = mgh/t = (60*10*8)/0.75 = 6400W
Practice Problems
• Pg 112 #2, 3, 4, 6, 10, 11, 13-16