Transcript Chapter 6: Work and Energy

Chapter 6: Work and Energy
Neither quantity is a vector!
Woo hoo!
Work and Energy
• In physics, work = a force applied over a
distance: W = Fd
• But if the force is not in the same direction,
then you only care about that part of the force
that is in the same direction
• So W = Fdcos(θ)
• In physics, work and energy are equivalent
• Both work and energy are measured in Joules
• 1 joule = 1 Newton•meter
Work: Sample prob 1
• Kody Clegg is in the weight room doing a
bench press. He lifts a total of 110 kg a
distance of 0.8 m. His force is in the same
direction the bar travels. What is the work
Kody does on the bar?
• W = (110kg)(9.8m/s2)(0.8m) = 862 Joules
• How much work is done as he lowers the bar?
• W = (110kg)(9.8m/s2)(-0.8m) = -862 Joules
• So what is meant by “negative work”?
Work: Sample prob 2
• Linda Monayong is pulling a piece of luggage.
She pulls upwards at an angle of 500 from the
horizontal with a force of 45 N and pulls the
luggage a total of 75 m (the luggage moves
horizontally along the floor). What is the work
she has done on the luggage?
• W = Fdcos(θ) = (45N)(75m)(cos(500))
• W = 2170 N
Work: Sample prob 3
• Some poor sap whose car (1000kg) won’t start is
trying to push it, but it’s stuck. He pushes with a
force of 200 N. What work is he doing on the car?
• 0 Joules (no distance moved)
• David Green is standing on top of a railroad
flatcar, arms crossed and looking studly. David’s
weight is 590 N. The boxcar travels 12 meters.
How much work does David’s weight do on the
boxcar as it travels?
• 0 Joules (force is 900 to direction of travel)
Kinetic Energy
• Kinetic Energy is the energy associated with
motion.
• If an object has a mass M and is moving at a
velocity V, then its kinetic energy is
• KE = ½ MV2
• Kinetic Energy is measured in Joules
Easy Kinetic Energy Sample Problems
• A 1100 kg car is traveling at 20 m/s. What is
the car’s kinetic energy?
• ½ * 1100 kg * (20 m/s)2 = 220,000 Joules
• A 50 gram bumblebee is flying at 0.75 m/s.
What is it’s kinetic energy?
• ½ * 0.05 kg * (0.75)2 = 0.014 J
• You get the idea
Work-Energy Theorem
• The work-energy theorem simply states that
the amount of work you do on a system
equals the change in kinetic energy.
• Work = ∆KE, where ∆KE = KEF – KEI, just like
always.
• If the mass of the object doesn’t change
(which is usually the case), then the change in
kinetic energy is due to a change in velocity
• ∆KE = KEF – KEI = ½mVF2 – ½mVI2
Work-Energy Problem
• A satellite of mass 5X104 kg is travelling at
speed V0 = 1.1X104 m/s. Its engine exerts a
force parallel with the direction of travel of a
magnitude 4X105 N and pushes over a
distance of 2.5X106 m. What is the final
velocity of the satellite?
• Work = Fd, so W = (4X105 N)(2.5X106 m) =
1X1012 J
• Work = ∆KE, so KEF = W - KEI
Work-Energy Problem continued
• Work = ∆KE, so KEF = W - KEI
• ½ (5 X 104kg)(VF2) = 1 X 1012 J –
½(5X104kg)(1.1X104m/s)2
• Solving for VF, we get VF = 1.27 X 104 m/s
Now, how did we solve that?
• First, we understand that the work done on a
system is equal to the change in kinetic
energy.
• Recall that Work = force * distance and solve
for work done
• Then, recall that KE = ½ mv2 and recognize
that ∆KE = KEF – KEI,
• We know enough info to solve for KEI
• All that is left is to solve for the unknown part
of KEF
Work – Energy concepts
• A satellite is in a circular orbit around the
earth and gravity is the only force acting on it.
• Does the gravity do any work?
• Does the satellite’s KE change?
Work-Energy concepts continued
• Now consider the case of the earth’s orbit
around the sun, which is actually an ellipse,
not a circle.
• Does the sun do any work on the earth?
Where? Where does it not do work?
• Why is this different
From a circular orbit?
http://www.oglethorpe.edu/faculty/~m_rulison
/Astronomy/Chap%2017/Images/ellipse.gif
Gravitational Potential Energy
• Force of gravity on an object (weight) is mg
• Work done by gravity is that force * a distance
• Let the distance over which weight acts be
height h
• So work done by gravity W = mgh
• Since work is equivalent to energy, then mgh
is equal to what we call the gravitational
potential energy
Dr. Mason, we are intrigued. Tell us
more about his “potential energy”, as
you call it.
• Potential energy is stored energy that has not
been used yet.
• Potential energy can be converted into other
kinds of energy (notably kinetic energy).
• The important thing about potential energy is
that it must be measured RELATIVE TO
something.
• Gravitational PE depends on height and you
have to measure the height from the ground.
• With no reference point, PE is meaningless
Falling stuff: Converting energy
• Let’s say a 3 kg rock is sitting at rest at a height of
10 meters above the ground. What is its speed
just before it hits the ground?
• PE = mgh, so PE = (3kg)(9.8 m/s2)(10m) = 294 J
• PE = ∆KE = KEF – KEI and KEI = 0 Joules (because
rock is initially at rest, so V = 0 m/s)
• So PE = 294 J = KEF = ½ mVF2
• 294 J = ½ (3kg)(VF2), so VF2 = 196 J, so VF= 14 m/s
So how does that compare to the old
way?
• How would we have found the speed of a
falling object before?
• Easiest way was Vf2 = Vi2 + 2AX and solve for Vf
• And actually our new method reduces to that
equation readily (do you see it?).
• Or we could have used F = ma, using a = Δv/Δt
and solving for time by using d = ½at2…
• You get the idea. In general, energy
conservation is an easier approach
Conservative and Nonconservative
Forces
• When talking about gravitational PE, we see
that all that matters is the height above the
ground.
• The object could fall straight down, or could
follow a diagonal path, or whatever. It doesn’t
matter. All that matters is the change in
height.
• A force is conservative if the work done by it
doesn’t depend on the path taken. Conversely,
a nonconservative force is one whose work
done does depend on the path taken.
Examples
• Conservative Forces
– Gravity
– Elastic springs
– Electrical force
• Nonconservative forces
– Friction
– Air resistance
– Tension
– Normal Force
Concepts
• Let’s say you have a conservative force, like
gravity. You fall from one height to the ground,
then get raised back up to the original height.
What is your net change in energy?
• Now consider a block sliding along a rough
surface for 1 m. If the block slides back to its
starting point, is the net change in energy = 0?
More Concepts
• Let’s say that you have a marble on a
frictionless U-shaped track. It is released from
a height H on one side. How high above the
bottom does the marble roll on the other
side?
• Now, add friction. Is the height reached
greater or less?
More stuff
• Sometimes, both conservative and
nonconservative forces act on a system (e.g. a
block can slide under gravity down an rough
incline).
• Remember that work and energy are
equivalent, so we would have:
• mgh + WNC = ∆KE, or gravitational PE + work
done by nonconservative force = change in KE
Conservation of Mechanical Energy
• We have introduced potential energy and
kinetic energy.
• Total mechanical energy = PE + KE
• E = mgh + ½ mv2
• If there are no nonconservative forces, then
mechanical energy is conserved.
• EF = EI
• mghF + ½ mvF2 = mghI + ½ mvI2
EF
EI
More Conservation of Mechanical
Energy
http://www.physicsclassroom.com/class/energy/u5l2b21.gif
So, Let’s say that in words
• The total energy stays constant.
• As PE decreases (because height decreases),
KE increases
• PE is being converted to KE
• As PE goes down, KE goes up
Energy in a pendulum
http://serc.carleton.edu/images/sp/library/uncertainty/diagram_conservation_energy_si.jpg
Energy of a Pendulum
• At position A, the
pendulum has only PE
• Between A and B, it has a
mix of PE and KE (with PE
decreasing and KE
increasing)
• At B, it has purely KE
• Between B and C, it has a
mix again (with KE
decreasing and PE
decreasing)
Sample Problem
• A motorcycle daredevil tries to leap across a
canyon by driving horizontally off a cliff at 38
m/s. Ignoring air resistance, with what speed
does the motorcycle hit the other side?
H0 = 70m
H1 = 70m
Sample Problem continued
•
•
•
•
mghF + ½ mvF2 = mghI + ½ mvI2
Every term has mass in it, so mass cancels out
ghF + ½ vF2 = ghI + ½ vI2
The only term that is unknown is VF, so solve
for that.
• VF = √ vI2 + 2g(h0 – hF)
• VF = √ (38m/s)2 + 2(9.8)(70m-35m)
• = 46.2m/s
Now, see how much easier that was
than before???
• Previously, we had to
– Find the length of time the cycle was in the air
using the free fall equation
– Use that info to find the final vertical speed
– Use the Pythagorean theorem to combine
horizontal and vertical speeds to get total speed.
Another Sample Problem
• Tarzan is standing on a tree branch 8 m above the
jungle floor. He is holding a vine that just happens
to be 8m long and swings down. Assuming no air
resistance and so forth, find his speed at the
bottom of the swing.
• Can simplify equation because final height = zero
and initial speed = zero
• Mgh = ½ mv2
• M(9.8m/s2)(8m) = ½ mv2
• V =√ 156.8 m2/s2
= 12.52 m/s
Power
• As you have seen in the last lab experiment,
power = energy/time
• Power also = work/time, since work and
energy are equivalent
• W/time = Fd/t, but d/t = speed. So power also
= Fv
• Units: 1 joule/sec = 1 watt
• 1000 watts = 1 kilowatt
• 1 horsepower (hp) = 745 watts
Power example
• An 1100 kg car, starting from rest, accelerates
for 5 sec. the magnitude of acceleration is a =
4.6m/s2. what was the average power
generated by the net force on the car?
• F = ma = (1100 kg)(4.6m/s2) = 5060 N
• Vavg = ½(vf – v0) and vf = v0 + at = 23 m/s
• Power = Fv, so P = (5060N)(11.5m/s) =
5.82X104 watts (78hp)