Transcript File

Work and Energy
Work
• Work means many things in everyday life.
• However, in Physics, work is defined one way.
• Work = the product of the magnitude of the
displacement times the component of the
force parallel to the displacement.
• W = F|| d , where F|| is the force parallel to the
displacement, d. (Remember the boat from
our last test)
The power of θ
• The force parallel to the displacement is the x
component of the force and is given by Fcosθ
• So, W = Fdcosθ
• When θ = 0, cosθ = 1 and all of the force goes
into work and W =Fd
Units of Work
• Force x distance is going to give us
Newtons x meters (Nm)
• This unit takes the name joule (J): 1J = 1Nm
– Named in honor of James Prescott Joule an
English brew master, who discovered the link
between mechanical energy and heat. (more on
that when we get to thermodynamics)
Other units of work
• Any units of force and distance will give you a
unit for work.
• One such unit is the erg. 1erg = 1dyne x cm
• In British units (which Americans use as well)
the unit is the foot-pound.
• 1J = 107erg = 0.7376ft*lb
Unemployed
• Will a force always create work?
• If not, under what circumstances will there be
no work done?
Example
• A 50kg crate is pulled 40m along a horizontal
floor by a constant force exerted by a person,
FP = 100N, which acts at a 37o angle. The floor
is rough and exerts a friction force Ffr = 50N.
Determine the work done by each force acting
on the crate, and the net work done on the
crate.
Solution
• We have 4 forces: push, friction, weight, and
normal.
• Weight and normal have a θ of 90 (weight is
actually -90 or 270).
• cos90 = 0, so Fdcosθ = 0. The weight and
normal forces do NO WORK.
• Wpush = Fpushdcosθ = 100N(40m)cos37 = 3200J
• Wfr = Ffrd cos180 = (50N)(40m)(-1) = -2000J
Solution part 2
• Wnet = Wg + WN + WP + Wfr =
0 + 0 + 3200J -2000J
= 1200J
• OR
• Wnet = Fnet,xd = (FPcosθ – Ffr)d
= (100N cos37 – 50N)(40m) = 1200J
Kinetic Energy
• Is the energy associated with an object in
motion
• KE = 1/2mv2
Kinetic Energy and Work
• Remember, W= Fd
• F = ma
• So W = mad
• d= ∆x
• So W = ma∆x
The Math
• Remember way back into chapter
2, vf2 = vi2 + 2a∆x
• Solve for a∆x, plug in to get,
W = 1/2mvf2 – 1/2mvi2
Kinetic Energy and Work Cont.
• …Therefore, W = ∆KE
Example
• A 145g baseball is thrown with a speed of
25m/s.
What is its kinetic energy?
How much work was done on the ball to make
it reach this speed, if it started from rest?
Solution
• KE = 1/2mv2 = ½(0.145kg)(25m/s)2 = 45J
• Because the ball started at rest it had 0J of
kinetic energy at the beginning, so the net
work done is equal to the kinetic energy, 45J.
Example
• How much work is required to accelerate a
1000kg car from 20m/s to 30m/s?
Solution
• W = KE2 – KE1
•
= 1/2mv22 – 1/2mv12
•
= ½(1000kg)(30m/s)2 – ½(1000kg)(20m/s)2
•
= 2.5 x 105 J
Potential Energy
• Is associated with an object that has
the potential to move because of its
location relative to some other
location.
Gravitational Potential Energy
• PEg = mgh, where m is the mass, g is gravity,
and h is the height of the object.
• But, the height above what?
Example
• A 0.1g pencil slowly rolls off of a 1m high desk
and onto the floor. What was the pencil’s
potential energy when it was:
• A. on the desk w.r.t the desk?
• B. on the desk w.r.t the floor?
• C. on the floor w.r.t the desk?
Solution
• First, we need kg. Why?
– 0.1g * (1kg/1000g) = 1.0 x 10-4kg
• A. PE = mgh = 1.0 x 10-4kg(9.8m/s2)(0m) = 0J
• B. PE = mgh = 1.0 x 10-4kg(9.8m/s2)(1m)
= 9.8 x 10-4 J
• C. PE = mgh =1.0 x 10-4kg(9.8m/s2)(-1m)
= -9.8 x 10-4 J
• Can PE be negative?
Elastic Potential Energy
• The potential energy stored in a
compressed or stretched object, like
a spring.
• The length of a spring when no
forces are acting on it is called the
relaxed length.
More on Springs
• The spring constant, k, refers to how “stiff”
the spring is. The higher the k the harder it is
to squeeze the spring.
• If a spring is hard to squeeze, then there is a
lot of energy in the spring when compressed.
• The distance the spring is squeezed or
stretched from its relaxed length we call x.
• You need more energy to increase x.
The Point
•PEelastic =
2
½kx
Example
• A spring has a spring constant of
200N/m and is compressed 10cm
from its relaxed length. What
potential energy is stored in the
spring?
Solution
• PEelastic = ½kx2
• PE = ½(200N/m)(0.1m)2 = 1J
• Unit Check: N/m *m2 = Nm = J
Roller coasters
• Roller coasters are powered by gravitational
potential energy.
• At the beginning of the ride, work is done on
the coaster to give it potential energy.
• Then work is done by the coaster to turn its
potential energy into kinetic energy (it falls
down)
More on coasters
• A coaster that falls to the bottom of the hill
has turned all of its potential energy into
kinetic.
• If the coaster goes back up another hill, it
loses kinetic energy (slows down) and turns it
back into potential energy (gets higher).
Ponder this
• Can a coaster climb a hill higher than the one
it started at? Explain.
• What happens to the energy of the coaster
when it is halfway up a hill? Halfway down?
Mechanical Energy
• As you should realize by now, potential and
kinetic energy are closely related.
• KE and PE combine to make a new quantity,
E, the total mechanical energy.
• E = KE + PE
Conservation of energy
• Mechanical energy is conserved = it is not
created or destroyed.
• ΔKE + ΔPE = 0
• KE2 + PE2 = KE1 + PE1
• E2 = E1
Example: Falling rock
• If the original height of a stone is 3.0m,
calculate the stone’s speed when it has fallen
to 1.0m above the ground.
solution
• v1 = 0m/s h1 = 3.0m h2 = 1.0m g = 9.8m/s2
• 1/2mv12 + mgh1 = 1/2mv22 + mgh2
• 0 + m(9.8m/s2)(3.0m) =
½(m)(v22) + m(9.8m/s2)(1.0m)
• We can cancel all of the m’s and solve for v22
• v22 = 2[(9.8m/s2)(3.0m) –
(9.8m/s2)(1.0m)]=39.2m2/s2
• v2 = 6.3 m/s
Power
• In everyday life, when we say that is powerful
we usually mean that it can put out a lot of
force.
• In physics, the official definition of power is
this: the rate at which work is done

work W
P  average _ power 

time
t
The Units of Power
• Work/time would give us J/s
• A J/s has a short name, the Watt, in honor of
James Watt, the inventor of the steam engine.
• 1J/s = 1W
• Yes, this is the Watts like in light bulbs.
Horse Power
• Like all American units this one has a really
dumb origin story.
• James Watt needed a way to explain how
powerful his new steam engine was.
• He ran an experiment and concluded that a
“good horse” can work all day at a rate of
360ft*lb/s. He multiplied this by 1.5 just to
make the horse look better and got 550ft*lb/s
Horse Power
• So, Watt declared 1 horse power (1hp) to be
equal to 550ft*lb
• Putting this into SI units gives us:
1hp = 746W
Example
• A 70kg jogger runs up a long flight of stairs in
4.0s. The vertical height of the stairs is 4.5m.
• A.) Find the jogger’s power output in both
Watts and horsepower.
• B.) How much energy did this take?
Solution
• The work done is against gravity, so W = mgy.
• P = W/t = mgy/t = (70kg)(9.8m/s2)(4.5m) / 4.0s
= 770W
• Because 1 hp = 746W the person’s power out
put is just over 1hp.
• The energy required is E = Pt (because work is
energy transfer).
• E = (770J/s)(4.0s) = 3100J
Firing Mah Lazer
• The Nova laser at Lawrence Livermore
National Lab has 10 beams, each beam has a
power output greater than all of the power
plants in the United States. Where does this
power come from?
Power in terms of velocity
• It is often useful to write power in terms of v
instead of work. Here’s how:
• P = W/t = Fd/t (d/t = v) so

• P=Fv
• Note: this is the average velocity, d/t
Example
• Calculate the power required of a 1400kg car
under the following circumstances:
• A. the car climbs a 10 degree hill at 80km/hr
• B. the car accelerates along a flat road from
90km/hr to 110km/hr in 6.0s.
• Assume Ffr = 700N