Transcript The Work-Energy Theorem

Work Work is calculated by
multiplying the force applied by
the distance the object moves
while the force is being applied.
W = Fs
The unit of work is the
Newton•meter, also
called the joule (J). The
equation we will use is:
W = (F cosϴ)s
Ex. 1 - Find the work done
by a 45.0 N force in
pulling a luggage carrier
at an angle ϴ = 50.0° for
a distance s = 75.0 m.
Ex. 2 - A weight lifter is bench
pressing a 710 N barbell. He
raises the barbell 0.65 m above
his chest and then lowers it the
same distance. What work is
done on the barbell during the
lifting and lowering phase.
Ex. 4 - A flatbed truck
2
accelerating at a = +1.5 m/s
is carrying a 120 kg crate. The
crate does not slip as the truck
moves s = 65 m. What is the
total work done on the crate
by all the forces acting on it?
Work done to an object
results in a change in the
kinetic energy of the object.
This relationship is called the
work-energy theorem.
W = Fs = mas
2
2
vf = v0 +2as can be solved
for as.
2
2
as = 1/2 (vf - v0 ), this
second term is substituted in
the first equation.
2
2
Fs = 1/2 m vf - 1/2 m v0
Fs = 1/2 m vf2 - 1/2 m v02
Work equals final kinetic energy
minus initial kinetic energy.
KE = 1/2 mv2
The unit is the joule.
The Work-Energy
Theorem.
W = KEf - KE0 =
1/2
2
mvf
- 1/2
2
mv0
Ex. 5 - A space probe of mass
m = 5.00 x 104 kg is traveling at a
speed of v0 = 1.10 x 104 m/s through
deep space. The engine exerts a
constant external force of
F = 4.00 x 105 N, directed parallel to
the displacement. The engine fires
continually during the displacement of
s = 2.50 x 106 m. Determine the final
speed of the probe.
Ex. 6 - A 54 kg skier is coasting
down a 25° slope. A kinetic
frictional force of fk = 70 N opposes
her motion. Her initial speed is
v0 = 3.6 m/s. Ignoring air
resistance, determine the speed vf
at a displacement 57 m downhill.
The work-energy
theorem deals with the
work done by the net
external force, not an
individual force (unless
its the only one).
The gravitational force is
a force that can do
positive or negative work.
W = (mg cos ϴ°)(h0 - hf) =
mg(h0 - hf)
Ex. 8 - A gymnast leaves a
trampoline at a height of 1.20 m
and reaches a height of 4.80 m
before falling back down.
Determine (a) the initial speed v0
with which the gymnast leaves
the trampoline and (b) the speed
of the gymnast after falling back
to a height of 3.5 m.
Gravitational Potential
Energy is energy due to
the distance an object is
able to fall.
PE = mgh
PE is also measured in joules.
Wgravity = mgh0 - mghf
= PE0 - PEf
The work done by the
gravitational force on an
object does not depend on
the path taken by the object.
This makes gravitational
force a conservative force.
•A force is conservative when the
work it does on a moving object is
independent of its path.
•A force is conservative when it
does no net work on an object
moving around a closed path,
starting and finishing at the same
point.
Nonconservative forces
are those where the work
done does depend on the
path. Kinetic frictional
forces and air resistance
are two examples.
In the work-energy
theorem both
conservative and
nonconservative forces
act on an object.
So: W = Wc + Wnc.
If work done is equal to
the change in KE and
Wc is due to
gravitational force,
then
Wnc = (KEf – KE0) + (PEf – PE0)
Conservation of Mechanical Energy
Wnc = (KEf - KE0) + (PEf - PE0)
becomes:
Wnc = (KEf + PEf) - (KE0 + PE0)
or:
Wnc = Ef - E0
Conservation of
Mechanical Energy
The total mechanical energy
(E = KE + PE) of an object remains
constant as the object moves,
provided that the net work done
by external nonconservative forces
is zero.
Ex. 9 - A motorcyclist drives
horizontally off a cliff to leap across
a canyon. When he drives off, he
has a speed of 38.0 m/s. Find the
speed with which the cycle strikes
the ground on the other side if he is
35 m below his starting point when
he strikes the ground.
Ex. 10 - A 6.00-m rope is tied to
a tree limb and used as a swing.
A person starts from rest with
the rope held in a horizontal
orientation. Determine how fast
the person is moving at the
lowest point on the circular arc
of the swing.
Ex. 12 - The roller coaster
Magnum XL-200 includes a
vertical drop of 59.4 m.
Assume that the coaster has
a speed of nearly zero as it
crests the top of the hill. Find
the speed of the riders at the
bottom of the hill.