Transcript work power energy - White Plains Public Schools

Work and Energy
Unit 4
Lesson 1 : Work Done by a Constant Force
When a force acts on an object while
displacement occurs, the force has done
work on the object.
The magnitude of work (W) is the product
of the amount of the force applied along
the direction of displacement and the
magnitude of the displacement.
W = Fcosq Dx
Units of Work
N . m = Joule (J)
Determining the Sign of Work
+ Work
If the force has a
component in the
direction of the
displacement.
- Work
If the force has a
component in the
opposite direction of
the displacement.
Example 1
Rank the following situations in order of
the work done by the force on the
object, from most positive to most
negative. [Displacement is to the right
and of the same magnitude.]
Example 2
m = 0.30
30o
Find the work done by all forces as a 4.0 kg
mass slides 5.0 m down a 30o incline where
the coefficient of kinetic friction is 0.30.
Graphical Analysis of Work
F (N)
F
WF = FDx
Dx
x (m)
Work done is the area under the graph
Work is a Scalar (Dot) Product
Since
W = Fcosq Dx
and
A . B = ABcosq
then
W = F . Dx
Example 3
A particle moving in the xy plane undergoes a
displacement Dx = (2.0 ^i + 3.0 ^j) m as a
^
^
constant force F = (5.0 i + 2.0 j) N acts on the
particle.
a) Calculate the magnitudes of the
displacement and the force.
b) Calculate the work done by force F.
Lesson 2 : Work Done by a Varying Force
xf
W = SFxDx
xi
As Dx approaches 0,
xf
lim S FxDx =
Dx  0
 F dx
x
xi
Therefore,
xf
W=
 F dx
x
xi
Example 1
A force acting on a particle varies with
x, as shown above. Calculate the work
done by the force as the particle
moves from x = 0 to x = 6.0 m.
Example 2
The interplanetary probe shown above is
attracted to the Sun by a force given by
1.3 x 1022
F= x2
This equation is in SI units, where x is
the Sun-probe separation distance.
Determine how much work is done by
the Sun on the probe as the probe-Sun
separation changes from 1.5 x 1011 m to
2.3 x 1011 m.
Graphical Solution
~ 60 squares
Each square = (0.05 N)(0.1 x 1011 m)
= 5 x 108 J
Work Done by a Spring
Hooke’s Law
force exerted by
spring
Fs = -kx
spring constant
in N/m
position relative to
equilibrium position
Negative sign signifies that the force
exerted by spring is always directed
opposite to the displacement.
stretched spring
equilibrium position
compressed spring
xf
Ws =
 F dx =
s
xi
 (-kx)dx =
½ kx2
Ws = ½ kx2
Work done by the spring force is positive
because the force is in the same direction
as displacement.
Generalized Work Done by Spring
xf
Ws =

(-kx)dx = ½ kxi2 - ½ kxf2
xi
Generalized Work Done on Spring
xf
Ws =

xi
Fappdx =
xf

xi
kxdx = ½ kxf2 - ½ kxi2
Example 3
A common technique used to measure the spring
constant (k) is shown above. The spring is hung
vertically, and an object of mass m is attached to
its lower end. Under the action of the “load” mg,
the spring stretches a distance d from its
equilibrium position.
a) If a spring is stretched 2.0 cm by a suspended
mass of 0.55 kg, what is the spring
constant of the spring ?
b) How much work is done by the spring as it
stretches through this distance ?
c) Suppose the measurement is made on an
elevator with an upward vertical
acceleration a. Will the unaware
experimenter arrive at the same value of
the spring constant ?
Example 4
If it takes 4.00 J of work to stretch a Hooke’s
Law spring 10.0 cm from its unstressed
length, determine the extra work required to
stretch it an additional 10.0 cm.
Example 5
A light spring with spring constant 1200 N/m
is hung from an elevated support. From its
lower end a second light spring is hung,
which has spring constant 1800 N/m. An
object of mass 1.50 kg is hung at rest from the
lower end of the second spring.
a) Find the total extension distance of the
pair of springs.
b) Find the effective spring constant of the
pair of springs as a system. We
describe these springs as in series.
Example 6
^
^
A force F = (4xi + 3yj) N acts on an
object as the object moves in the xdirection from the origin to x = 5.00 m.

Find the work W = F . dx done on the
object by the force.
Lesson 3 : Work-Kinetic Energy Theorem
Work done by SF is
xf
SW =
 SF dx
xi
xf
 ma dx
SW =
xi
xf

SW =
dv
m
dx
dt
xi
xf

(by chain-rule) SW =
xi
vf
SW =
dv
m
dx
 mv
dx
dx
dt
dv
vi
SW = ½ mvf2 – ½ mvi2
Kinetic Energy
KE = ½ mv2
Work - Kinetic Energy Theorem
If work done on a system only
changes its speed, the work done
by the net force equals the
change in KE of the system.
SW = KEf – KEi = DKE
Example 1
A 6.0 kg block initially at rest is pulled to the
right along a horizontal, frictionless surface by
a constant horizontal force of 12 N. Find the
speed of the block after it has moved 3.0 m.
Example 2
A man wishes to load a refrigerator onto a
truck using a ramp. He claims that less
work would be required to load the truck if
the length L of the ramp were increased. Is
his statement valid ?
Example 3
A 4.00 kg particle is subject to a total force that
varies with position as shown above. The particle
starts from rest at x = 0. What is its speed at
a) x = 5.00 m
b) x = 10.00 m
c) x = 15.00 m
Lesson 4 : Situations Involving Kinetic Friction
SFx = max
(SFx)Dx = (max)Dx
ax =
vf - vi
t
Dx = ½ (vi + vf) t
(SFx)Dx = m
(
vf - vi
t
) ½ (v + v ) t
i
f
(SFx)Dx = ½ mvf2 – ½ mvi2
This is not work because Dx is displacement of
a particle – the book is not a particle !
(SFx)Dx = -fkDx = ½ mvf2 – ½ mvi2 = DKE
-fkDx = DKE
DKE in General
-fkd = DKE
d = length of any
path followed
DKE = -fkd + SWother forces
OR
KEf = KEi - fkd + SWother forces
Example 1
A 6.0 kg block initially at rest is pulled to
the right along a horizontal surface by a
constant horizontal force of 12 N.
a) Find the speed of the block after it has
moved 3.0 m if the surfaces in
contact have a coefficient of kinetic
friction of 0.15.
b) Suppose the force F is applied at and
angle q as shown below. At what angle
should the force be applied to achieve
the largest possible speed after the
block has moved 3.0 m to the right ?
Change in Internal Energy due to Friction
The result of a friction force is to
transform KE into internal energy, and
the increase in internal energy is equal
is equal to the decrease in KE.
DEsystem = DKE + DEint = 0
-fkd + DEint = 0
DEint = fkd
Example 2
A 40.0 kg box initially at rest is pushed 5.00
m along a rough, horizontal floor with a
constant applied horizontal force of 130 N. If
the coefficient of friction between box and
floor is 0.300, find
a) the work done by the applied force
b) the increase in internal energy in the
box-floor system due to friction
c) the work done by the normal force
d) the work done by the gravitational
force
e) the change in kinetic energy of the box
f) the final speed of the box.
Lesson 5 : Power
Same amount of work done
Time interval is different
Average Power
time rate of energy transfer
P=
W
Dt
Instantaneous Power
P = lim
Dt  0
P=
dW
dt
=
W
Dt
=
F . dx
dt
dW
dt
. v
F
=
Units of Power
SI unit of power is J/s or the Watt (W).
1 W = 1 J/s = 1 kg . m2/s3
In the U.S. customary system, the unit of
power is the horsepower (hp).
1 hp = 746 W
The kilowatt-hour (kWh)
The energy transferred in 1 h at the
constant rate of 1kW = 1000 J/s.
1 kWh = (103 W)(3600 s) = 3.60 x 106 J
* Note that a kWh is a unit of energy, not
power.
Example 1
An elevator car has a mass of 1600 kg and is
carrying passengers having a combined mass
of 200 kg. A constant friction force of 4000 N
retards its motion upward, as shown above.
a) What power delivered by the motor is
required to lift the elevator car at a
constant speed of 3.00 m/s ?
b) What power must the motor deliver at
the instant the speed of the elevator
is v if the motor is designed to
provide the elevator car with an
upward acceleration of 1.00 m/s2 ?
Example 2
Find the instantaneous power delivered by
gravity to a 4 kg mass 2 s after it has fallen
from rest.
Example 3
Find the instantaneous power delivered by the
net force at t = 2 s to a 0.5 kg mass moving in
one dimension according to x(t) = 1/3 t3.
Example 4
Consider a car of mass m that is accelerating up a
hill, as shown above. An automotive engineer
measures the magnitude of the total resistive force
to be ft = (218 + 0.70v2) N where v is in m/s.
Determine the power the engine must deliver to the
wheels as a function of speed.
Example 5 : AP 2003 #1
100 kg
The 100 kg box shown above is being
pulled along the x-axis by a student. The
box slides across a rough surface, and its
position x varies with time t according to
the equation x = 0.5t3 + 2t, where x is in
meters and t is in seconds.
a) Determine the speed of the box at time
t = 0.
b) Determine the following as functions of
time t.
i. The kinetic energy of the box.
ii. The net force acting on the box.
iii. The power being delivered to the box.
c) Calculate the net work done on the box
in the interval t = 0 to t = 2 s.
d) Indicate below whether the work done
on the box by the student in the
interval t = 0 to t = 2 s would be greater
than, less than, or equal to the answer
in part c). Justify your answer.
____Greater than
____ Less than
____ Equal to
Lesson 6 : Potential Energy
system = book + Earth
Work done on
system by external
agent in lifting
book
DKE = 0
(vi = 0, vf = 0)
When book is at yb, the energy of the
system has potential to become KE.
Gravitational Potential Energy
When lifting at constant velocity,
^.
^
W = (Fapp) Dx = (mgj) [(yb – ya)j] = mgyb - mgya
.
Ug = mgy
W = DUg
Units for Ug are Joules (J). Like work
and KE, Ug is a scalar quantity.
Example 1
A bowling ball held by a careless bowler
slips from the bowler’s hands and drops
on the bowler’s toe. Choosing floor level
as the y = 0 point of your coordinate
system, estimate the change in
gravitational PE of the ball-Earth system
as the ball falls. Repeat the calculation,
using the top of the bowler’s head as the
origin of coordinates.
Example 2
A 400 N child is in a swing that is attached
to ropes 2.00 m long. Find the
gravitational potential energy of the childEarth system relative to the child’s lowest
position when
a) the ropes are horizontal
b) the ropes make a 30o angle with the
vertical
c) the child is at the bottom of the circular
arc.
Lesson 7 : Conservation of Mechanical Energy
As book falls from yb to ya,
the work done by the
gravitational force on the
book is
^
^
Won book = (mg) (Dx) = (-mg j) . [(ya – yb) j]
.
Won book = mgyb - mgya
From the work-kinetic energy theorem,
Won book = DKEbook
DKEbook = mgyb - mgya
For the book-Earth system,
mgyb – mgya = -(mgya – mgyb) = -(Uf – Ui) = -DUg
So,
DKE = -DUg
Bringing DU to left side of the equation,
DKE + DUg = 0
This sum of KE and Ug is called
mechanical energy.
Emech = KE + U
represents all
types of potential
energy
(KEf – KEi) + (Uf – Ui) = 0
KEf + Uf = KEi + Ui
Conservation of Mechanical Energy
(isolated, frictionless system)
Elastic Potential Energy
WFapp = ½ kxf2 – ½ kxi2
Elastic potential energy
stored in a spring
Us = ½ kx2
Example 1
A ball of mass m is
dropped from a height h
above the ground, as
shown.
a) Neglecting air resistance,
determine the speed of
the ball when it is at a
height y above the
ground.
b) Determine the speed of the ball at y if at
the instant of release it already has an
initial upward speed vi at the initial
altitude h.
Example 2
A pendulum consists of a sphere of
mass m attached to a light cord of
length L. The sphere is released from
rest at point A when the cord makes an
angle qA with the vertical, and the pivot
at P is frictionless.
a) Find the speed of the sphere when it is
at the lowest point B.
b) What is the tension TB in the cord at B ?
Example 3
The launching mechanism of a
toy gun consists of a spring of
unknown spring constant.
When the spring is compressed
0.120 m, the gun, when fired
vertically, is able to launch a
35.0 g projectile to a maximum
height of 20.0 m above the
position of the projectile before
firing.
a) Neglecting all resistive forces,
determine the spring constant.
b) Find the speed of the projectile as it
moves through the equilibrium position
of the spring (where xB = 0.120 m).
Example 4
A bead slides without friction around a loopthe-loop. The bead is released from a height
h = 3.50R.
a) What is its speed at point A ?
b) How large is the normal force on it if
its mass is 5.00 g ?
Example 5
An object of mass m starts from rest and slides
a distance d down a frictionless incline of angle
q. While sliding, it contacts an unstressed
spring of negligible mass as shown above. The
object slides an additional distance x as it is
brought momentarily to rest by compression of
the spring (of spring constant k). Find the initial
separation d between object and spring.
Example 6 : AP 1989 # 1
A 0.1 kg block is released from rest at point A as
shown above, a vertical distance h above the
ground. It slides down an inclined track, around a
circular loop of radius 0.5 m, then up another
incline that forms an angle of 30o with the
horizontal. The block slides off the track with a
speed of 4 m/s at point C, which is a height of 0.5 m
above the ground. Assume the entire track to be
frictionless and air resistance to be negligible.
a) Determine the height h.
b) On the figure below, draw and label all
the forces acting on the block when
it is at point B, which is 0.5 m above
the ground.
c) Determine the magnitude of the force
exerted by the track on the block
when it is at point B.
d) Determine the maximum height above
the ground attained by the block
after it leaves the track.
e) Another track that has the same
configuration, but is NOT frictionless, is
used. With this track it is found that if
the block is to reach point C with a
speed of 4 m/s, the height h must be 2 m.
Determine the work done by the frictional
force.
Example 7 : AP 1985 # 2
An apparatus to determine coefficients of friction
is shown above. The box is slowly rotated
counter-clockwise. When the box makes an angle
q with the horizontal, the block of mass m just
starts to slide, and at this instant the box is
stopped from rotating. Thus at angle q, the block
slides a distance d, hits the spring of force
constant k, and compresses the spring a distance
x before coming to rest.
In terms of the given quantities, derive an
expression for each of the following.
a) ms, the coefficient of static friction
b) DE, the loss in total mechanical energy of
the block-spring system from the start
of the block down the incline to the
moment at which it comes to rest on
the compressed spring
c) mk, the coefficient of kinetic friction
Lesson 8 : Conservative and Nonconservative Forces
Conservative Forces
1. The work done by a conservative force
on a particle moving between any two
points is independent of the path
taken by the particle.
2. The work done by a conservative force
on a particle moving through any
closed path is zero. (A closed path is
one in which the beginning and end
points are identical.)
Examples of Conservative Forces
a) Gravitational Force
Wg = mgyi - mgyf
yi
Fg
Wg depends on y
coordinates and is
independent of the path
yf
Wg is zero when the object
moves over any closed
path (where yi = yf).
Fg
b) Force exerted by a spring
Ws = ½ kxi2 – ½ kxf2
Ws depends on y
coordinates and is
independent of the path
Wg is zero when the object
moves over any closed
path (where yi = yf).
Nonconservative Forces
A force that does not satisfy the
properties of a conservative force.
Work done by force depends on the path.
Nonconservative forces
acting within a system
cause a change in the
mechanical energy of the
system.
If book is displaced along blue path, work
done against friction is less than if book is
pushed along curved brown path.
Friction force is a nonconservative force.
If the forces acting on objects within a system
are conservative, then the mechanical energy
of the system is conserved.
If some of the forces acting on objects within
a system are nonconservative, then the
mechanical energy of the system changes.
If a friction force acts within a system,
DEmech = DKE + DU = -fkd
Example 1
A 3.00 kg crate slides down a ramp. The
ramp is 1.00 m in length and inclined at an
angle of 30.0o. The crate starts from rest at
the top, experiences a constant friction
force of magnitude 5.00 N, and continues
to move a short distance on the horizontal
floor after it leaves the ramp. Use energy
methods to determine the speed of the
crate at the bottom of the ramp.
Diagram for Example 1
Example 2
A child of mass m rides
on an irregularly curved
slide of height h = 2.00 m.
The child starts from rest
at the top.
a) Determine his speed at the bottom,
assuming no friction is present.
b) If a force of kinetic friction acts on the
child, how much mechanical energy
does the system lose ? Assume that
vf = 3.00 m/s and m = 20.0 kg.
Example 3
Two blocks are connected by a light string that passes over
a frictionless pulley. The block of mass m1 lies on a
horizontal surface and is connected to a spring of force
constant k. The system is released from rest when the
spring is unstretched. If the hanging block of mass m2 falls
a distance h before coming to rest, calculate the coefficient
of kinetic friction between the block m1 and the surface.
Lesson 9 : Conservative Forces and PE
The work done by a conservative
force equals the decrease in PE of
the system.
xf
Wc =
F
x
dx = -DU
xi
xf
DU = Uf – Ui = -
F
x
dx
xi
DU = is negative when Fx and dx are
in the same direction.
F
F
equiibrium
DUg is negative
DUs is negative
xf
Uf (x) = -
F
xi
x
dx + Ui
dU = -Fx dx
dU
Fx = dx
The x-component of a conservative force
acting on an object within a system equals
the negative derivative of the potential
energy of the system with respect to x.
Gravitational PE
dUg
Fg = dy
d
Fg = dy
Fg = -mg
(mgy)
Elastic PE
dUs
Fs = dx
d
Fs = (1/2 kx2)
dx
Fs = -kx
(Hooke’s Law)
Example 1
Consider the potential energy of two
molecules given by
U=
A
r12
B
r6
Find the force along the line joining
the two molecules.
Lesson 10 : Energy Diagrams
KE
PE
Negative slope
equals F
“Stable” equilibrium
U(x) is a minimum
The force at a given point is the
negative slope of the curve.
F=
dU
= -kx
dx
Where the graph reaches maxima or
minima, the force will be 0.
Stable equilibrium points will be
located at the minima.
Fx is negative
Fx is positive
Acceleration
away from x = 0
Acceleration
away from x = 0
“unstable” equilibrium
Example 1
For the potential energy curve shown below,
a) determine whether the force Fx is positive,
negative, or zero at the five points
indicated.
b) indicate points of stable, unstable, and
neutral equilibrium.
c) sketch the curve for Fx vs. x from x = 0 to
x = 9.5 m
Example 2
A particle moves along a line where the
potential energy of its system depends on its
position r as graphed below. In the limit as r
increases without bound, U(r) approaches +1J.
a) Identify each equilibrium position for this
particle. Indicate whether each is a point
of stable, unstable, or neutral
equilibrium.
b) The particle will be bound if the total
energy of the system is in what range ?
Now suppose that the system has energy -3J.
Determine
c) the range of positions where the particle
can be found.
d) its maximum kinetic energy.
e) the location where it has maximum kinetic
energy.
f) the binding energy of the system – that is,
the additional energy that it would have
to be given in order for the particle to
move out to r  infinity .
Example 3
Jane, whose mass is 50.0 kg, needs to swing across a river
(having width D) filled with man-eating crocodiles to save
Tarzan from danger. She must swing into a wind exerting
constant horizontal force F, on a vine having length L and
initially making an angle q with the vertical.
Taking D = 50.0 m, F = 110 N, L = 40.0 m,
and q = 50.0o,
a) with what minimum speed must Jane
begin her swing in order to just make it
to the other side ?
b) Once the rescue is complete, Tarzan and
Jane must swing back across the river.
With what minimum speed must they
begin their swing ? Assume that Tarzan
has a mass of 80.0 kg.
Example 4 : AP 1987 # 2
The following graph shows the potential energy U(x)
of a particle as a function of its position x.
a) Identify all points of equilibrium for this
particle.
Suppose the particle has a constant total
energy of 4.0 J, as shown by the dashed line
on the graph.
b) Determine the kinetic energy of the
particle at the following positions :
i. x = 2.0 m
ii. x = 4.0 m
c) Can the particle reach the position
x = 0.5 m ? Explain.
d) Can the particle reach the position
x = 5.0 m ? Explain.
e) On the grid below, carefully draw a graph
of the conservative force acting on the
particle as a function of x, for 0<x<7 m.
Example 5 : AP 1995 # 2
A particle of mass m moves in a conservative
force field described by the potential energy
function U(r) = a(r/b + b/r), where a and b are
positive constants and r is the distance from the
origin. The graph of U(r) has the following shape.
a) In terms of the constants a and b,
determine the following :
i. The position ro at which the potential
energy is a minimum.
ii. The minimum potential energy Uo.
b) Sketch the net force on the particle as a function
of r on the graph below, considering a force
directed away from the origin to be positive,
and a force directed toward the origin to be
negative.
The particle is released from rest at r = ro/2.
c) In terms of Uo and m, determine the speed
of the particle when it is at r = ro.
d) Write the equation or equations that could
be used to determine where, if ever, the
particle will again come to rest. It is not
necessary to solve for this position.
e) Briefly and qualitatively describe the
motion of the particle over a long period
of time.