Transcript CHAPTER 7) WORK AND KINETIC ENERGY 7.1) Work Done by a Constant Force • What is work? Consider Figure (7.1) F F F Figure 7.1 (a) (b) (c) • A.

CHAPTER 7) WORK AND KINETIC ENERGY
7.1) Work Done by a Constant Force
• What is work? Consider Figure (7.1)
F
F
F
Figure 7.1
(a)
(b)
(c)
• A force is applied to an object, and the object slides along the horizontal
direction.
• If we are interested in how effective the force is in moving the object, we
consider (i) the magnitude of the force, and (ii) its direction.
• Assume that the magnitude of the applied force is the same in all three cases.
• The push applied in Figure (7.1b) does more to move the object than the push in
Figure (7.1a).
• Figure (7.1c) – the applied force does not move the eraser at all, regardless of
how hard it is pushed.
•
In analyzing forces to determine the work required (to cause that motion)
–
consider (i) the vector nature of forces, and (ii) how far the object moves along
the horizontal direction.
•
Example – moving the eraser 3 m requires more work than moving it 2 cm.
•
Figure (7.2) – an object undergoes a displacement d along a straight line while
acted on by a constant force F that makes an angle  with d.
•
The work W done on an object by an agent exerting a constant force on the
object is the product of the component of the force in the direction of the
displacement and the magnitude of the displacement:
W  Fd cos 
(7.1)
The situation where a force does no work (W=0)
1.
If the object does not move – i.e if d=0, Equation (7.1) gives W = 0 (the
situation in Figure (7.1c)). Example – holding a chair.
2.
When the force applied is perpendicular to the object’s displacement – i.e. if
 = 90o, then from Eq. (7.1), W = 0 because cos 90o = 0. Example: Figure
(7.3) – the work done by the normal force on the object and the work done by
the force of gravity on the object = zero because both forces are perpendicular
to the displacement and have zero components in the direction of d.
The sign of the work done by the applied force
• Depends on the direction of F relative to d.
• Positive when the vector associated with the component F cos  is in the
same direction as the displacement.
Example – when an object is lifted, the work done by the applied force is
positive because the direction of that force is upward, the same direction as
the displacement.
• Negative when the vector associated with the component F cos  is in the
direction opposite the displacement.
Example – as an object is lifted, the work done by the gravitational force on
the object is negative.
• The factor cos  in the definition of W (Eq. (7.1)) automatically takes care of
the sign.
• Work is an energy transfer; if energy is transferred to the system (object), W is
positive; if energy is transferred from the system, W is negative.
• If an applied force F acts along the direction of the displacement, then  = 0
and cos 0o = 1.
• In this case, Equation (7.1) gives :
W = Fd
• Work is a scalar quantity, and its units are force multiplied by length.
• The SI unit of work = newton·meter (N·m) = joule (J).
• In general, a particle may be moving with either a constant or a varying
velocity under the influence of several forces.
• In these cases, because work is a scalar quantity, the total work done as the
particle undergoes some displacement is the algebraic sum of the amounts of
work done by all the forces.
Example (7.1) : Mr. Clean
A man cleaning a floor pulls a vacuum cleaner with a force of magnitude
F
= 50.0 N at an angle of 30.0o with the horizontal (Fig. (7.4a)). Calculate the
work done by the force on the vacuum cleaner as the vacuum cleaner is
displaced 3.00 m to the right.
7.2) The Scalar Product of Two Vectors
Scalar Product
• Indicate how F and d interact in a way that depends on how close to parallel
they happen to be.
• Scalar product = F·d (Also known as the dot product – because of the dot
symbol).
• Equation (7.1) becomes :
W = F·d = Fd cos 
(7.2)
• F·d (read “F dot d”) is a shorthand notation for Fd cos .
• In general, the scalar product of any two vectros A and B is a scalar quantity
equal to the product of the magnitudes of the two vectors and the cosine of the
angle  between them :
A ·B  AB cos 
(7.3)
• A and B need not have the same units.
• Figure (7.6) – B cos  is the projection of B onto A. Equation (7.3) sayas that
A ·B is the product of the magnitude of A and the projection of B onto A.
• The scalar product is commutative :
A ·B = B ·A
That is, the order of the dot product can be reversed.
• The scalar product obeys the distributive law of multiplication :
A ·( B + C ) = A ·B + A ·C
• If A is perpendicular to B ( = 90o), then A ·B = 0.
• The case of A ·B =0, when either A or B is zero.
• If vector A is parallel to vector B and the two point in the same direction
( = 0o) , then A ·B = AB.
• If vector A is parallel to vector B but the two point in opposite directions
( = 180o), then A ·B = - AB.
• The scalar product is negative when 90o <  < 180o.
• The unit vector i, j, and k – lie in the positiove x, y, and z directions,
respectively, of a right-handed coordinate system.
• It follows from the definition of A ·B that the scalar products of these unit
vectors are :
i ·i = j ·j = k ·k = 1
(7.4)
i ·j = i ·k = j ·k = 0
(7.5)
• Two vectors A and B can be expressed in component vector form as :
A = Ax i + Ay j + Az k
B = Bx i + By j + Bz k
• Using the information given in Equations (7.4) and (7.5) – the scalar product of
A and B reduces to :
A ·B = Ax Bx + Ay By + Az Bz
• In special case in which A = B :
A  A  A2x  A2y  A2z  A2
(7.6)
Example (7.2) : The Scalar Product
The vector A and B are given by A =2i + 3j and B =-i + 2j. (a) Determine the
scalar product A ·B.
Example (7.3) : Work Done by a Constant Force
A particle moving in the xy plane undergoes a displacement d = (2.0i + 3.0j) m
as a constant force F = (5.0i + 2.0j) N acts on the particle. (a) Calculate the
magnitude of the displacement and that of the force.
7.3) Work Done by a Varying Force
• Consider a particle being displaced along the x axis under the action of a
varying force.
• The particle is displaced in the direction of increasing x from x = xi to x = xf.
• In such a situation, we cannot use W = (F cos ) d to calculate the work done
by the force because this relationship applies only when F is constant in
magnitude and direction.
• If we imagine that the particle undergoes a very small displacement x
(Figure (7.7a)) – the x component of the force Fx is approximately constant
over this interval.
• For this small displacement, we can express the work done by the force as :
W  Fx x
The area of the shaded rectangle in Fig. 7.7a
• Imagine that the Fx versus x curve is divided into a large number of such
intervals.
• Then the total work done for the displacement from xi to xf is approximatelay
equal to the sum of a large number of such terms :
xf
W   Fx x
xi
• If the displacements are allowed to approach zero, then the number of terms in
the sum increases without limit but the value of the sum approaches a definite
value equal to the area bounded by the Fx curve and the x axis :
xf
xf
xi
xi
lim  Fx x   Fx dx
x 0
• This definite integral is numerically equal to the area under the Fx-versus-x
curve between xi and xf.
• Therefore, we can express the work done by Fx as the particle moves from xi to
xf
xf as :
(7.7)
Work done by a varying force
W   Fx dx
xi
• If more than one force acts on a particle, the total work done is just the work
done by the resultant force.
• If we express the resultant force in the x direction as Fx , then the total work,
or net work, done as the particle moves from xi to xf is :
xf
 W  Wnet    Fx dx
xi
(7.8)
Example (7.4) : Calculating Total Work Done from a Graph
A force acting on a particle varies with x, as shown in Figure (7.8). Calculate
the work done by the force as the particle moves from x = 0 to x = 6.0 m.
Fx(N)
A
5
0
B
Figure (7.8)
C
1
2 3
4 5 6
x(m)
Example : A block of mass 2.5 kg is pushed 2.2 m along a frictionless
horizontal table by a constant 16.0-N force directed 25o below the horizontal.
Determine the work done by (a) the applied force, (b) the normal force exerted
by the table, (c) the force of gravity, and (d) the net force on the block.
Example : A force F = (6i – 2j) N acts on a particle that undergoes a
displacement s = (3i + j) m. Find (a) the work done by the force on the
particle and (b) the angle between F and s.
Example : The force acting on a particle varies as in Figure (P7.24). Find the
work done by the force as the particle moves (a) from x = 0 to x = 8.0 m,
(b) from x = 8.0 m to x = 10 m, and (c) from x = 0 to x = 10 m.
Fx (N)
B
6
A
0
C
2
4
6
8
E
10
-2
D
12
x (m)
Work Done by a Spring
• Figure (7.10) – A common physical system for which the force varies with
position.
• A block on a horizontal, frictionless surface is connected to a spring.
• If the spring is either stretched of compressed a small distance from its
unstretched (equilibrium) configuration, it exerts on the block a force of
magnitude :
(7.9)
Spring force
Fs  kx
where x = the displacement of the block from its unstretched (x=0) position
k = a positive constant called the force constant of the spring.
• The force required to stretch or compress a spring is proportional to the amount
of stretch or compression x.
• This force law for springs = Hooke’s law (valid only in the limiting case of
small displacements).
• The value of k is a measure of the stiffness of the spring. Stiff springs have
large k values, and soft springs have small k values.
• The negative sign (Eq. (7.9)) – signifies that the force exerted by the spring is
always directed opposite the displacement.
• When x > 0 (Figure (7.10a)) – the spring force is directed to the left, in the
negative x direction.
• When x < 0 (Figure (7.10c)) – the spring force is directed to the right, in the
positive x direction.
• When x = 0 (Figure (7.10c)) – the spring is unstretched and Fs = 0.
• Because the spring force always acts toward the equilibrium position (x = 0), it
sometimes is called a restoring force.
• If the spring is compressed until the block is at the point -xmax and then
released  the block moves from -xmax through zero to +xmax.
• If the spring is stretched until the block is at the point xmax and is then
released  the block moves from +xmax through zero to -xmax.
• It then reverses direction, returns to +xmax , and continues oscillating back and
forth.
Suppose the block has been pushed to the left a distance xmax from equilibrium
and is then released
• Calculate the work Ws done by the spring force as the block moves from
xi = -xmax to xf = 0.
• Apply Equation (7.7), and assume the block may be treated as a particle :
xf
0
Ws   Fs dx   (kx)dx  12 kx2max
xi
(7.10)
 x max
• The work done by the spring force is positive because the force is in the same
direction as the displacement (both are to the right).
• Consider the work done by the spring force as the block moves from xi = 0 to
xf = xmax.
Ws   12 kx2max because the displacement is to the right and
• We find that
the spring force is to the left.
• Therefore, the net work done by the spring force as the block moves from
xi = - xmax to xf = xmax is zero.
• Figure (7.10d) – a plot of Fs versus x.
• The work calculate in Eq. (7.10) is the area of the shaded triangle,
corresponding to the displacement from -xmax to 0.
• Because the triangle has base xmax and height kxmax, its area is ½ kx2max , the
work done by the spring as given by Eq. (7.10).
• If the block undergoes an arbitrary displacement from x=xi to x=xf , the work
done by the spring force is :
xf
Ws   (kx)dx  12 kxi2  12 kxf2
(7.11)
Work done by a spring
xi
• From Eq. (7.11), the work done by the spring force is zero (Ws = 0) – for any
motion that ends where it began (xi = xf).
Work done on the spring by an external agent
• The external agent stretches the spring very slowly form xi = 0 to xf = xmax
(Figure (7.11)).
• Calculate the work by noting that at any value of the displacement, the
applied force Fapp is equal to and opposite the spring force Fs , so that
Fapp = - (-kx) = kx.
• The work done by this applied force (the ecternal agent) is :
x max
x max
0
0
WFapp   Fappdx   kxdx  12 kx2max
This work is equal to the negative of the work done by
the spring force for this displacement
Fs
Fapp
Figure (7.11)
xi = 0
xf = xmax
Example (7.6) : Measuring k for a Spring
• A common technique used to measure the force
constant of a spring is described in Figure
(7.12).
• The spring is hung vertically, and an object of
mass m is attached to its lower end.
mg (0.55kg)(9.80m / s 2 )
k

d
2.0 10 2 m
 2.7 102 N / m
• Under the action of the “load” mg, the spring
stretches a distance d from its equilibrium
position.
• Because the spring force is upward (opposite
the displacement) it must balance the
downward force of gravity mg when the
system is at rest.
Fs
d
(a)
• In this case, we apply Hooke’s law to give
|Fs| = kd = mg , or
k
mg
d
• For example, if a spring is stretched 2.0 cm by
a suspended object having a mass of 0.55 kg,
then the force constant is :
(b)
mg
Figure (7.12)
(c)
7.4) Kinetic Energy andThe Work-Kinetic Energy Theorem
• Figure (7.13) - a particle of mass m moving to the right under the action of a
constant net force F.
• Because the force is constant - from Newton’s second law, the particle moves
with a constant acceleration a.
• If the particle is displaced a distance d, the net work done by the total force F
is :
(7.12)
 W  Fd  (ma)d
• The relationships below are valid when a particle undergoes constant
acceleration :
d
d  12 ( v i  v f ) t
v  vi
a f
t
m
F
Figure (7.13)
vi
where vi = the speed at t = 0 and vf = the speed at time t.
vf
• Substituting these expressions into Equation (7.12) :
 vf  vi  1
W

m

 2 v i  v f t

 t 
 W  12 mvf2  12 mvi2
(7.13)
2
• The quantity 12 mv represents the energy associated with the motion of the
particle = kinetic energy.
• The net work done on a particle by a constant net force F acting on it equals
the change in kinetic energy of the particle.
• In general, the kinetic energy K of a particle of mass m moving with a speed v
is defined as :
(7.14)
K  12 mv2
• Kinetic energy is a scalar quantity and has the same units as work = Joule (J).
• It is often convenient to write Equation (7.13) in the form :
W  K
• That is :
f
 Ki  K
Ki   W  K f
(7.15)
Work-kinetic energy theorem
Work-kinetic energy theorem
• Must include all of the forces that do work on the particle in the calculation of
the net work done.
• The speed of a particle increases if the net work done on it is positive - because
the final kinetic energy is greater than the initial kinetic energy.
• The speed of a particle decreases if the net work done is negative because the
final kinetic energy is less than the initial kinetic energy.
• Kinetic energy is the work a particle can do in coming to rest, or the amount of
energy stored in the particle.
• We derived the wrok-kinetic energy theorem under the assumption of a constant
net force - but it also is valid when the force varies.
• Suppose the net force acting on a particle in the x direction is Fx.
• Apply Newton’s second law, Fx = max, and use Equation (7.8), the net work
done :
xf
xf
 W    F dx   ma
x
xi
xi
x
dx
• If the resultant force varies with x, the acceleration and speeed also depend on x.
• Acceleration is expressed as :
a
dv dv dx
dv

v
dt dx dt
dx
• Substituting this expression for a into the above equation for W gives :
xf
v
f
dv
 W   mv dx dx   mvdv
xi
vi
2
2
1
1
W

mv

mv
 2 f 2 i
(7.16)
The net work done on a
particle by the net force
acting on it is equal to the
change in its kinetic
energy of the particle
• Situation Involving Kinetic Friction
• Analyze the motion of an object sliding on a horizontal surface - to describe the
kinetic energy lost because of friction.
• Suppose a book moving on a horizontal surface is given an initial horizontal
velocity vi and slides a distance d before reaching a final velocity vf
(Figure (7.15)).
d
vi
Figure (7.15)
vf
fk
• The external force that causes the book to undergo an acceleration in the
negative x direction is the force of kinetic friction fk acting to the left, opposite
the motion.
• The initial kinetic energy of the book is ½mvi2 , and its final kinetic energy is
½mvf2 (Apply the Newton’s second law).
• Because the only force acting on the book in the x direction is the friction force,
Newton’s second law gives -fk = max.
• Multiplying both sides of this expression by d and using Eq. (2.12) in the form
vxf2 - vxi2 = 2axd
for motion under constant acceleration give
-fkd = (max)d = ½mvxf2 - ½mvxi2 or :
(7.17a)
Kfriction  f k d
• The amount by which the force of kinetic friction changes the kinetic energy of
the book is equal to -fkd.
• Part of this lost kinetic energy goes into warming up the book, and the rest
goes into warming up the surface over which the book slides.
• The quantity -fkd is equal to the work done by kinetic friction on the book
plus the work done by kinetic friction on the surface.
• When friction (as welll as other forces) - acts on an object, the work-kinetic
energy theorem reads :
(7.17b)
Ki  Wother  f k d  Kf

• Wother represents the sum of the amounts of work done on the object by
forces other than kinetic friction.
Example (7.7) : A Block Pulled on a Frictionless Surface
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.
n
vf
F
mg
n
fk
d
F
mg
(a)
vf
d
(b)
Figure (7.16)
Example (7.8) : A Block Pulled on a Rough Surface
Find the final speed of the block described in Example (7.7) if the surface is not
frictionless but instead has a coefficient of kinetic friction of 0.15.
Example (7.9) : Does the Ramp Lessen the Work Required?
A man wishes to load a refrigerator onto a truck using a ramp, as shown in
Figure (7.17). 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?
Figure (7.17)
L
Example (7.10) : Useful Physics for Safer Driving
A certain car traveling at an initial speed v slides a distance d to a halt after its
brakes lock. Assuming that the car’s initial speed is instead 2v at the moment
the brakes lock, estimate the distance it slides.
Example (7.11) : A Block-Spring System
A block of mass 1.6 kg is attached to a horizontal spring that has a force constant
of 1.0x103 N/m (Figure (7.10)). The spring is compressed 2.0 cm and is then
released from rest. (a) Calculate the speed of the block as it passes through the
equilibrium position x = 0 if the surface is frictionless.
7.5) Power
• The two cars - has different engines (8 cylinder powerplant and 4 cylinder
engine), same mass.
• Both cars climb a roadway up a hill, but the car with the 8 cylinder engine
takes much less time to reach the top.
• Both cars have done the same amount of work against gravity, but in
different time periods.
• Know (i) the work done by the vehicles, and (ii) the rate at which it is done.
• The time rate of doing work = power.
• If an external force is applied to an object (assume acts as a particle), and if the
work done by this force in the time interval t is W, then the average power
expended during this interval is defined as :

W
t
Average power
• The work done on the object contributes to the increase in the energy of the
object.
• Therefore, a more general definition of power is the time rate of energy
transfer.
• We can define the instantaneous power P as the limiting value of the average
power as t approaches zero :
W dW
  lim

t  0 t
dt
Represent increment of work done by dW
• From Equation (7.2), letting the displacement be ecpressed as ds, that
dW = F·ds.
• Therefore, the instantaneous power can be written :
dW
ds

 F  Fv
dt
dt
(7.18)
v =ds / dt
• The SI unit of power = joules per second (J /s) = watt (W).
1W = 1 J/s = 1 kg ·m2 / s3
• A unit of power in the British engineering system is the horsepower (hp) :
1 hp = 746 W
• A unit of energy (or work) can now be defined in terms of the unit of power.
• One kilowatt hour (kWh) = the energy converted or consumed in 1 h at the
constant rate of 1 kW = 1 000 J /s.
1kWh = (103W)(3600 s) = 3.60 x 106 J
kWh = a unit of energy, not power
Example (7.12) : Power Delivered by an Elevator Motor
An elevator car has a mass of 1 000 kg and is carrying passengers having a
combined mass of 800 kg. A constant frictional force of 4 000 N retards its
motion upward, as shown in Figure (7.18a). (a) What must be the minimum
power delivered by the motor to lift the elevator car at a constant speed of
3.00 m/s?