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Chapter 4
Chapter 4
Energy and Work
Energy and Work
Basic Requirements
1. Understand the concept of work and kinetic energy;
2. Master the work-kinetic energy theorem;
3. Master the work done by a gravitational force and by a spring force；
4. Master the work done by a general variable force;
5. Understand the concept of power.
6. Understand the concept of potential energy;
7. Understand the path Independence of conservative forces;
8. Master the law of conservation of mechanical energy;
9. Master the work done on a system by an external force;
10. Master the law of conservation of energy.
Review and Summary
Kinetic Energy The kinetic energy K associated with motion of a particle of mass
m and speed v , where v is well below the speed of light, is
K
1 2
mv
2
(kinetic energy)
(4-1)
Work Work W is energy transferred to or from an object via a force acting on the
object. Energy transferred to the object is positive work, and energy transferred from
the object is negative work.
Work Done by a Constant Force The work done on a particle by a constant force
F during displacement d of the particle is
W  Fd cos   F  d (work, constant force)
(4-7,4-8)
in which  is the constant angle between the direction of F and d . Only the
component of F that is along the displacement d can do work on the object. When
Chapter 4
Energy and Work
two or more forces act on an object, their net work is the sum of the individual
works by the forces, which is also equal to the work that would be done on the object
by the net force Fnet of those forces.
Work and kinetic energy We can relate a change K in kinetic energy of a particle
to the net work W done on the particle with
K  K f  K i  W (work-kinetic energy theorem)
(4-10)
in which K i is the initial kinetic energy of the object and K f is the kinetic energy
after the work is done.
Work Done by the Gravitational Force The work W g done by the gravitational
force Fg on a particle –like object of mass m during a displacement d of the object
is given by
Wg  mgd cos 
(4-12)
in which  is the angle between Fg and d .
Work Done in Lifting and Lowering an Object The work Wa done by an applied
force during a lifting or lowering of a particle-like object is related to the work
W g done by the gravitational force and the change K in the object’s kinetic energy
by
K  K f  K i  Wa  Wg
(4-15)
If the kinetic energy at the beginning of a lift equals that at the end of the lift, then Eq.
reduces to
Wa  Wg
(4-16)
which tells that the applied force transfers as much energy to the object as the
gravitational force transfers from the object.
Spring Force The force F from a spring is
Chapter 4
Energy and Work
F   kd (Hooke’s law)
(4-18)
where d is the displacement of its free end from the position when the spring is in its
relaxed state (neither compressed nor extended), and k is the spring constant (a
measure of the spring’s stiffness). If an x axis lies along the spring, with the origin at
the location of the spring’s free end when the spring is in its relaxed state, Eq. can be
written as
F  kx (Hooke’s law)
(4-19)
A spring force is thus a variable force: It varies with the displacement of the spring’s
free end.
Work Done by a Spring Force If an object is attached to the spring’s free end, the
work Ws done on the object by the spring force when the object is moved from an
initial position xi to a final position x f is
Ws 
1 2 1 2
kxi  kx f
2
2
(4-23)
If xi  0 and x f  x , then Eq. becomes
Ws 
1 2
kx
2
(4-24)
Work Done by a Variable Force When the force F on a particle-like object
depends on the position of the object, the work done by F on the object while the
object moves from an initial position ri with coordinates ( xi , yi , zi ) to a final
position r f with coordinates ( x f , y f , z f ) must be found by integrating the force. If
we assume that component Fx may depend on x but not on y or z , component
Fy may depend on y but not on x or z , and component Fz may depend on z but
not on y or x , then the work is
xf
yf
zf
xi
yi
zi
W   Fx dx   Fy dy   Fz dz
(4-34)
Chapter 4
Energy and Work
If F has only an x component, then Eq. (4-34) reduces to
xf
W   Fx dx
xi
(4-35)
Power The power due to a force is the rate at which that forces does work on an
object. If he force does work W during a time interval t , the average power due to
the force over that time interval is
Pavg 
W
t
(4-40)
Instantaneous power is the instantaneous rate of doing work:
P
dW
dt
(4-41)
If the direction of a force F is at an angle  to the direction of travel of the object,
the instantaneous power is
P  Fv cos   F  v
(4-45, 4-46)
in which v is the instantaneous velocity of the object.
conservative force A force is a conservative force if the net work it does on a
particle moving around every closed path, from an initial point and then back to that
point, is zero. Equivalently, it is conservative if the net work it does on a particle
moving between two points does not depend on the path taken by the particle. The
gravitational force and the spring force are conservative forces; the kinetic frictional
force is a-nonconservative force.
Potential energy A potential energy is energy that is associated with the
configuration of a system in which a conservative force acts. When the conservative
force does work W on a particle within the system, the change U in the potential
energy of the system is
U  W
(4-47)
If the particle moves from point xi to point x f , the change in the potential energy
of the system is
xf
U    F ( x)dx
xi
Gravitational Potential Energy
(4-52)
The potential energy associated with a system
Chapter 4
Energy and Work
consisting of Earth and a nearby particle is gravitational potential energy. If
the particle moves from height yi to height y f , the change in the gravitational
potential energy of the particle-Earth system is
U  mg ( y f  yi )  mg y
(4-53)
If the reference position of the particle is set as yi  0 and the corresponding
gravitational potential energy of the system is set as U i  0 , then the gravitational
potential energy U when the particle is at any height y is
U ( y )  mgy
(4-55)
Elastic Potential Energy Elastic potential energy is the energy associated with the
state of compression or extension of an elastic object. For a spring the exerts a spring
force F  kx when its free end has displacement x , the elastic potential energy is
U ( x) 
1 2
kx
2
(4-57)
The reference configuration has the spring at its relaxed length, at which x  0 and
U 0.
Mechanical Energy The mechanical energy Emec of a system is the sum of its
kinetic energy K and its potential energy U :
Emec  K  U
(4-58)
An isolated system is one in which no external force causes energy changes. If only
conservative forces do work within an isolated system, then the mechanical energy
Emec of the system cannot change. This principle of conservation of mechanical
energy is written as
K2  U 2  K1  U1
(4-63)
in which the subscripts refer to different instants during an energy transfer process.
This conservation principle can also be written as
Chapter 4
Energy and Work
Emec  K  U  0
(4-64)
Potential Energy Curves If we know the potential energy function U ( x) for a
system in which a one-dimensional force F acts on a particle, we can find the force
as
F ( x)  
dU ( x)
dx
(4-68)
If U ( x) is given on a graph, then at any value of x , the force F is the negative of
the slope of the curve there and the kinetic energy of the particle is given by
K ( x)  Emec  U ( x)
(4-70)
where Emec is the mechanical energy of the system. A turning point is a point x at
which the particle reverses its motion (there, K  0 ). The particle is in equilibrium at
points where the slope of the U ( x) curve is zero (there, F ( x)  0 ).
Work Done on a System by an External Force Work W is energy transferred to or
from a system by means of an external force acting on the system. When more than
one force acts on a system, their net work is the transferred energy. When friction is
not involved, the work done on the system and the change Emec in the mechanical
energy of the system are equal:
W  Emec  K  U
(4-71, 4-72)
When a kinetic friction force acts within the system, then the thermal energy Eth of
the system changes. (This energy is associated with the random motion of atoms and
molecules in the system.) The work on the system is then
W  Emec  Eth
(4-79)
The change Eth is related to the magnitude f k of the frictional force and the
magnitude d of the displacement caused by the external force by
Chapter 4
Energy and Work
Eth  f k d
(4-77)
Conservation of Energy The total energy E of a system (the sum of its mechanical
energy and its internal energies, including thermal energy) can change only by
amounts of energy that are transferred to or from the system. This experimental facts is
known as the law of conservation of energy. If work W is done on the system, then
W  E  Emec  Eth  Eint
(4-80)
If the system is isolated ( W  0 ), this gives
and
Emec  Eth  Eint  0
(4-81)
Emec ,2  Emec ,1  Eth  Eint
(4-82)
where the subscripts 1 and 2 refer to two different instants.
Power The power due to a force is the rate at which that force transfers energy. If an
amount of energy E is transferred by a force in an amount of time t , the average
power of the force is
Pavg 
E
t
(4-85)
The instantaneous power due to a force is
P
dE
dt
(4-86)
Examples
Example 1 As shown in Fig. 4-1, a skier skies downhill from rest from point A of a
hilltop of 50m and the length of the slope is 500m . When the skier reaches the
bottom of point B she continues until she reaches point C where the length of the
horizontal track that the skier goes through ? Near point B the trace can be
considered as smoothly varying and ignore the effect of air resistance.
Solution: Consider the skier, the track, and the Earth as a system, since the effect of
the air resistance on the skier is ignored, the forces on the skier are the gravity P , the
support force FN , and the frictional force Fr , where the gravity is a conservative
internal force, and only the non-conservative internal force, i.e., the frictional force,
Chapter 4
Energy and Work
does work, no other external forces do work. We then know from the principle of
work and energy that I the process of the skier skiing the frictional force does the work
as follows:
W  W1  W2  ( E p 2  Ek 2 )  ( E p1  Ek1 )
Fig. 4-1 Example 1
where W1 and W2 are the work done by the frictional force on the skier when she
slides downhill and skies on the flat track respectively. E p1 and Ek 1 are the potential
energy and the kinetic energy of the skier at the hilltop, E p 2 and Ek 2 are the
potential energy and the kinetic energy of the skier at rest at the end of the track. If we
choose the potential energy of the skier on the horizontal track to be zero, from the
question we know that E p1  mgh , E p 2  0 , E p 2  0 and Ek 2  0 , then from
the above equation we have
W1  W2  mgh
Also, from the definition of work, we have
B
B
A
A
W1   F  dr   Ff dr    mg cos dr
Chapter 4
Energy and Work
Since the gradient of the slope is very small, we have
W1    mgs
W2   F  dr    mgs
Solving for s , we have
s
h

 s
As given by the question, h  50m,   0.050, s  500m , we obtain the length
that the skier goes through on the horizontal track:
s
50
m  500m  500m
0.050
(Answer)
We should point out that this problem can also be solved by applying Newton’s
second law, i.e., obtain the acceleration first and then apply the formula of the
uniformly accelerated motion on a straight line, but the calculation will be much more
complicated and the readers can find it out by themselves.
Example 2
(a) Determine the work a hiker must do on a 15.0kg backpack to carry
it up a hill of height h  10.0m , as shown in Fig. 4-2a. Determine also (b) the work
done by gravity on the backpack, and (c) the net work done on the backpack. For
simplicity, assume the motion is smooth and at constant velocity (i.e., there is
negligible acceleration).
Solution: (a) The forces on the backpack are shown in Fig. 4-2b: the force of gravity,
mg , acting downward; and FH , the force the hiker must exert upward to support the
pack. Since we assume there is negligible acceleration, the net horizontal force is
negligible. In the vertical direction, we choose up as positive. Newton’s second law
applied to the backpack gives
F
y
 may , FH  mg  0
Hence, FH  mg  (15.0kg )(9.8m / s )  147 N
2
Chapter 4
Energy and Work
To calculate the work done by the hiker on the backpack, Eq. 4-8 can be written
WH  FH (d cos  )
and we note from Fig. that d cos  h . So the work done by the hiker can be
written:
WH  FH (d cos  )  FH h  mgh  147 N 10.0m  1470J
(Answer)
Fig. 4-2 Example 2
Note that the work done depends only on the change in elevation and not on the angle
of the hill,  . The same work would be done to lift the pack vertically the same
height h .
(b) The work done by gravity is :
WG  ( FG )(d ) cos(180   )
Since cos(180   )   cos  , we have
WG  ( FG )(d )( cos  )  mg (d cos  )  mgh
 (15.0kg )(9.8m / s 2 )(10.0m)  1470 J
(Answer)
Note that the work done by gravity doesn’t depend on the angle of the incline but only
on the vertical height h of the hill. This is because gravity does work only in the
vertical direction. We will make use of this important result later.
(c) The net work done on the backpack is Wnet  0 , since the net force on the
backpack is zero. We can also determine the net work done by writing
Wnet  WG  WH  1470 J  1470 J  0
(Answer)
Chapter 4
Energy and Work
which is, as it is should be , the same result.
Note in this example that even thought the net work on the backpack is zero, the
hiker nonetheless does do work on the backpack equal to 1470J.
Example 3 A horizontal spring has constant k  360N / m . (a) How much work is
required to compress it from its uncompressed length ( x  0 ) to x  11.0cm ? (b) If
a 1.85kg block is placed against the spring and the spring is released, what will be
the speed of the block when it separates from the spring at x  0 ? Ignore friction. (c)
Repeat part (b) but assume that the block is moving on a table as in Fig. 4-3 and that
the coefficient of kinetic friction is
k  0.38 .
Solution: (a) We saw W , needed to stretch or compress a spring by a distance x is
1 2
kx . Therefore the work required to compress the spring a distance
2
x  0.110m is
W
Fig. 4-3 Example 3
1
W  (360 N / m)(0.110m) 2  2.18 J
2
(Answer)
(b) In returning to its uncompressed length, the spring does 2.18J of work on the block.
According to the work-energy principle, the block acquires kinetic energy of the 2.18J.
Since K 
1 2
mv , the block’s speed must be
2
Chapter 4
v
Energy and Work
2K
2(2.18 J )

 1.54m / s
m
1.85kg
(Answer)
(c) There are two forces on the block: that exerted by the spring and that by
friction. The spring does 2.18J work on the block. Since the normal force FN
equals the weight mg the work done by the friction force on the block,
k FN  k mg , is
Wff  (k mg )( x)  (0.38)(1.85kg )(9.8m / s 2 )(0.110m)  0.76 J
This work is negative because the force of friction is in the direction opposite to the
displacement x . The net work done on the block is
Wnet  2.18J  0.76 J  1.42 J .
From the work-energy principle, we have
v
2(1.42 J )
 1.24m / s
1.85kg
(Answer)
Example 4 A ball of mass m  2.60kg , starting from rest, falls a vertical distance
h  55.0cm before striking a vertical coiled spring, which it compresses (see Fig. 4-4)
an amount Y  15.0cm . Determine the spring constant of the spring. Assume the
spring has negligible mass. Measure all distance from the point where the ball first
touches the uncompressed spring ( y  0 at this point).
Solution: Since the motion is vertical , we use y instead of x . We divide this
solution into two parts. Part 1: Let us first consider the energy changes of the ball as it
falls from a height y1  h  0.55m, Fig.4-4a, to y2  0 , just as it touches the spring,
Fig. 4-4b. Our system is the ball acted on by gravity (so far, the spring does nothing)
so
1 2
1
mv1  mgy1  mv22  mgy2
2
2
Chapter 4
Energy and Work
0  mgh 
1 2
mv2  0
2
and
v2  2 gh  2(9.8m / s 2 )(0.550m)  3.28m / s
Part 2: As the ball compresses the spring, Fig.4-4b to c, there are two conservative
forces on the ball-gravity and the spring force. So our energy equation becomes
E(ball touches spring)=E(spring compressed)
1 2
1
1
1
mv2  mgy2  ky22  mv32  mgy3  ky32
2
2
2
2
Fig. 4-4 Example 4
We take point 2 to be the instant when the ball just touches the spring, so y2  0 and
v2  3.28m / s . Point 3 we take to be when the ball comes to rest and the spring is
fully compressed, so v3  0 and y3  Y  0.150m .Putting these into the above
energy equation we get
1 2
1
mv2  0  0  0  mgY  kY 2
2
2
We know m, v2 and Y , so we can solve for k
Chapter 4
Energy and Work
2 1 2
m
[ mv2  mgY ]  2 [v22  2 gY ]
2
Y 2
Y
(Answer)
2.60kg
2
2

[(3.28m / s) 2(9.80m / s )(0.150m)]  1580 N / m
(0.150m) 2
k
Alternation solution: Instead of dividing the solution into two, we can do it at once.
After all, we get to choose what two points are used on the left and right of the energy
equation. Let us write the energy equation for points 1 and 3. Point 1 is the initial point
just before the ball starts to fall, so v1  0, y1  h  0.550m ; and point 3 is when the
spring is fully compressed, so v3  0, y3  Y  0.150m . The forces on the ball in
this process are gravity and the spring. So conservation of energy tells us
1 2
1
1
1
mv1  mgy1  k (0) 2  mv32  mgy3  ky32
2
2
2
2
1 2
0  mgh  0  0  mgY  kY
2
where we have set y  0 for the spring at point 1 because it is not acting and is not
compressed or stretched at this point. We solve for k :
2mg (h  Y ) 2(2.60kg )(9.8m / s 2 )(0.550m  0.150m)
k

 1580 N / m
Y2
(0.150m)2
just as in our first method of solution.
Example 5 The simple pendulum shown in Fig. 4-5 consists of a small bob of mass
m suspended by a massless cord of length l . The bob is released at t  0 , where the
cord makes an angle
   0 to the vertical. (a) Describe the motion of the bob in
terms of kinetic energy and potential energy. Then determine the speed of the bob: (b)
as a function of position  as it swings back and forth, and (c) at the lowest point of
the swing. (c) Find the tension in the cord, FT . Ignore friction and air resistance.
Solution: (a) At the moment of release, the bob is at rest, so K  0 . As the bob falls,
it loses potential energy is a maximum and the potential energy is a minimum. The bob
Chapter 4
Energy and Work
continues its swing until it reaches an equal height and angle on the opposite
side , at which point the potential energy is a maximum and K  0 . It continues the
swinging motion as U  K  U and so on, but it can never go higher than
  0 .
(b) The cord is assumed to be massless, so we don’t need to be concerned with the
energy of the cord but only with the bob’s kinetic energy, and the potential energy. The
bob has two forces acting on it at any moment gravity, mg , and the force the cord
exerts on it, FT . The latter always acts perpendicular to the motion, so it does no
work. We need be concerned only with gravity, for which we can write th potential
energy. The mechanical energy of the system is
E
1 2
mv  mgy
2
where y is the vertical height of the bob at any moment. We take y  0 at the
lowest point of the bob’s swing. Hence at t  0
y  y0  l  l cos 0  l (1  cos 0 )
Fig. 4-5 Example 5
as can be seen from the diagram. At the moment of release
E  mgy0
since v  v0  0 . At any other point along the swing
E
1 2
mv  mgy  mgy0
2
Chapter 4
Energy and Work
we solve this for v :
v  2 g ( y0  y )
 of the cord, we can write
In terms of the angle
v  2 gl (cos   cos 0 )
(Answer)
since y  l  l cos  and y0  l  l cos 0 .
(c) At the lowest point, y  0 , so
v  2 gy0 or v  2 gl (1  cos 0 )
(Answer)
(d) The tension in the cord is the force FT that the cord exerts on the bob. As we ‘ve
seen, there is no work done by this force, but we calculate the force simply by using
Newton’s second law
 F  ma
and by noting that any point the acceleration of
2
the bob in the inward radial direction is v / l , since the bob is constrained to move in
an arc of a circle. In the radial direction, FT acts inward, and a component of gravity
equal to mg cos  acts outward. Hence
v2
m  FT  mg cos 
l
2
We solve for FT and use the result of part (b) for v :
v2
FT  m(  g cos  )  2mg (cos   cos  0 )  mg cos 
l
 (3cos   2 cos  0 )mg
Example 6
(Answer)
Calculate the power required of a 1400kg car under the following
circumstances: (a) the car climbs a 10
0
hill at a steady 80km / h ; and (b) the car
Chapter 4
Energy and Work
accelerates along a level road from 90 to 110km / h in 6.0s to pass another
car. Assume the retarding force on the car is FR  700 N throughout. See Fig. 4-6.
(Be careful not to confuse FR , which is due to air resistance and friction that retard
the motion, with the force F needed to accelerate the car, which is the friction force
exerted by the road on the tires- the reaction to the motor-driven tires pushing against
the road.)
Solution: (a) To move at a steady speed up the hill, the car must exert a force equal to
the sum of the retarding force, 700N , and the component of gravity parallel to the
hill
mg sin10  (1400kg )(9.8m / s 2 )(0.174)  2400 N .
Since v  80km / h  22m / s and is parallel to F , then
P  Fv  (2400 N  700 N )(22m / s)  6.80 104W  91hp
(Answer)
Fig. 4-6 Example 6
(b) The car accelerates form 25.0m / s to 30.6m / s . Thus the car must exert a force
that overcomes the 700N retarding force plus that required to give it the acceleration
ax  (30.6m / s  25.0m / s) / 6.0s  0.93m / s 2
We apply Newton’s second law with x being the direction of motion:
max   Fx  F  FR
Chapter 4
Energy and Work
Then the force required, F , is
F  max  FR  (1400kg )(0.93m / s 2 )  700 N  2000 N
Since P  F  v , the required power increases with speed and the motor must be able
to provide a maximum power output of
P  (2000 N )(30.6m / s)  6.12 104W  82hp
Even taking into account the fact that only 60 to 80 percent of the engine’s power
output reaches the wheels, it is clear from these calculations that an engine of 100 to
150hp is more than adequate from a practical point of view.
Problem Solving
1 (2) Figure.4-7 shows an overhead view of three horizontal forces acting on a cargo
canister that was initially stationary but now moves across a frictionless floor. The
force magnitudes are F1  3.00 N , F2  4.00 N , F3  10.0 N , and the indicated
angles are  2  50.0 , 3  35.0 . What is the net work done on the canister by the
three forces during the first 4.00m of displacement?
Solution: The key idea here is that since the three forces are constant in magnitude
and direction, we can find the net force acting on the
cargo canister. First we calculate the net forces along
the x and the y axes:
Fx  F3 cos 3  F1  F2 sin 2
Fy  F3 sin 3  F2 cos 2
Substituting known values, yields
Fx  2.13N , Fy  3.17 N
So the net force can be obtained as
Fig. 4-7 Problem 1
Fnet  Fx2  Fy2  3.82 N
During the first 4.00m of displacement, the net work done on the canister is
Wnet  Fnet S  3.82 N  4.00m  15.28J
(Answer)
Chapter 4
Energy and Work
2 (3) In Fig.4-8, a block of ice slides down a frictionless ramp at angle   50
while an ice worker pulls on the block (via a rope) with a force Fr that has a
magnitude of 50.0N and is directed up the ramp. As the block slides through distance
d  0.5m along the ramp, its kinetic energy increases by 80 J . How much greater
would its kinetic energy have been if the rope had not been attached to the block?
Solution: Since the increasing kinetic energy is the sum of
the works done by the forces acting on the ice, we have:
K  WG  WFr  G sin 50 d  Fr d  80J
G sin 50 d  80 J  50  0.5 J  105 J
Thus, if the rope had not been attached to the block, the
increasing kinetic energy would have been:
Fig. 4-8 Problem 2
K '  G sin 50 d  105 J
(Answer)
3 (10) A fully loaded, slow-moving freight elevator has a cab with a total mass of
1200kg , which is required to travel upward 54m in 3.0 min , starting and ending
at rest. The elevator’s counterweight has a mass of only 950kg , and so the elevator
motor must help. What average power is required of the force the motor exerts on the
cab via the cable?
Solution: The key idea is that the power required is an average power. Since the
elevator’s counterweight has a mass of 950kg , the force exerted by the motor must
be
F  (m  m' ) g  (1200kg  950kg )  9.8N / kg  2450 N
The work done by the force can be yielded as
W  Fh  2450N  54m  132300J
So the average is required of the force the motor exerts on the cab via the cable is
P
W 132300 J

 740W
t
180s
(Answer)
4 (13) A 250 g block is dropped onto a relaxed vertical spring that has a spring
Chapter 4
Energy and Work
constant of k  2.5N / cm (Fig.4-9).The block becomes attached to the spring
and compress the spring 12cm before momentarily stopping. While the spring is
being compressed, what work is done on the block by (a) the gravitational force on it
and
(b) the spring force? (c) What is the speed of the block just before it hits the
spring?(Assume that friction is negligible.) (d) If the speed at impact is doubled, what
is the maximum compression of spring?
Solution: The key idea is that the spring will get the elastic potential energy of the
spring increases.
When the spring is being compressed, the direction of the
gravitational force is the same as the block’s moving, but the
direction of the elastic force is in the opposite direction. So the
work done by the gravitational force is positive while the spring
force does negative work on the block.
(a) Since the spring is compressed 12cm , the work done by the
gravitational force
Fig. 4-9 Problem 4
Wg  mgh  250kg  9.8N / m2  0.12m  294J
(Answer)
(b) The work done by the spring force can be given as
WS 
1 2 1
kx   250 N / m  (0.12m) 2  180 J
2
2
(Answer)
(c) Utilizing the Work-Kinetic Energy Theorem, the block’s speed can be obtained as
Wg  Ws 
1 2 1 2
mv f  mvi
2
2
here, the final speed v f is zero, solving for vi and substituting known values, we
find
vi  3.47m / s
(Answer)
(d) If the speed at impact is doubled, we use the Work-Kinetic Energy Theorem again
1
1
1
mgx  kx 2  mv 2f  mvi2
2
2
2
in which x is the maximum compression of the spring, substituting known values,
we get
x  23cm
(Answer)
5 (18) In Fig.4-10a, a block of mass m lies on a horizontal frictionless surface and is
Chapter 4
Energy and Work
attached to one end of a horizontal spring (spring constant k ) whose other end
is fixed. The block is initially at rest at the position where the spring is unstretched
( x =0) when the resulting kinetic energy of the block versus its position x is shown

in Fig.4-10.(a) What is the magnitude of F ? (b)What is the value of k ?
Solution: From work-kinetic theorem, we can
write
1
Fx1  kx12  K x 1  K x 0
2
1 2
Fx2  kx2  K x 2  K x 0
2
Substituting known values from the graph, we can find
Fig. 4-10 Problem 5
F  8 N , k  8 N / m2
(Answer)
6 (20) An iceboat is at rest on a frictionless frozen when a sudden wind exerts a
constant force of 200 N , toward the east, on the boat. Due to the angle of the sail, the
wind causes the boat to slide in a straight line for a distance of 8.0m in a direction
20 north of east. What is the kinetic energy of the iceboat at the end of 8.0m ?
Solution: The key idea is that the direction of the sail isn’t clear. Since we all know
that the boat’s shape is just like a shuttle, so the force’s component whose direction is
vertical to the profile, can only be effective. Thus, the direction of the sail is clear, it is
in the direction of 20 northwest, as shown in the figure on the right.
So the effective force acting on the boat is
F  200  cos 20  187.94 N
then, we use work-kinetic energy theorem, we can write
Fs  K f  K i
The initial kinetic energy K i is zero, and we can find the kinetic energy of the iceboat
at the end of 8.0m as
K f  Fs  187.94 N  8.0m  1504 J
(Answer)
Chapter 4
Energy and Work
7 (25) In Fig.4-11, a small block of mass
m  0.032kg
can slide along the frictionless
loop-the-loop, with loop radius R  12cm . The block
is released from rest at point P , at height h  5.0R
above the bottom of the loop. How much work does
the gravitational force do on the block travels from
point P to (a) point Q and (b) the top of the loop?
Fig. 4-11 Problem 7
If the gravitational potential energy of the block-Earth system is taken to be zero at the
bottom of the loop, what is that potential energy when the block is (c) at point P (d)
at point Q , and (e)at the top of the loop? (f) If, instead of merely being released, the
block is given some initial speed downward along the track, do the answers to (a)
through (e) increase, decrease, or remain the same?
Solution: (a) The work done by the gravitational force on the block as the block
travels from point P to Q is
Wg  mg (h  R)
Solving this equation, we can get
Wg  0 . 1 5 3J 6
(Answer)
(b) From the point P to the top of the loop, the work done by the gravitational force
is
Wg  mg (h  2 R)  0.1152 J
(Answer)
(c) We know the gravitational potential energy of the block-Earth system is taken to be
zero at the bottom of the loop, so we assume U P , U Q , U  are the potential energy
when the block is at point P, Q and at the top of the loop, respectively.
U P  m g h0 . 0 3 2k g 9 . 8m2 /s 5 0 . 1m2
(d)
0 . 1J 9 2
UQ  mgR  0.032kg  9.8m / s 2  0.12m  0.0384J
(Answer)
(Answer)
Chapter 4
(e)
Energy and Work
U   mg 2R  0.032kg  9.8m / s 2  2  0.12m  0.0768J
(Answer)
(f) The gravitational potential energy has nothing to do with the initial
speed of the block, so the answers to (a) through (e) remain the same.
8 (30) Figure 4-12 shows a pendulum of length L  1.25m . Its bob (which
effectively has all the mass) has speed v0 when the cord makes an angle  0  40.0
with the vertical. (a) What is the speed of the bob when it is in its
lowest position if v0  8.00m / s ? What is the least value that
v0 can have if the pendulum is to swing down and then up (b) to
a horizontal position, and (c) to a vertical position with the cord
remaining straight? (d) Do the answers to (b) and (c) increase,
decrease, or remain the same if  0 is increased by a few
degrees?
Solution: (a) The key idea is that the total energy of the system
cannot change. We can apply the law of conservation of energy in
the form of Emec  Eth  0 to this system. So we get
Fig. 4-12 Problem 8
1
1
mv2  mv02  mgl (1  cos  )  0
2
2
Substituting the known values, yields
1
1
m  v2  m  (8m / s ) 2  m  9.8m / s 1.25(1  cos 40 )  0
2
2
Solving this equation gives us
v  7.7m / s
(Answer)
(b) According to the law of conservation of energy, we have
1 2
m v0  m gclo s
2
Substituting the known values and solving this equation yields
v0  4.5m / s
(Answer)
(c) When the pendulum up to a vertical position with the cord remaining straight, we
can obtain
mv 2
 mg
L
Chapter 4
Energy and Work
Utilizing the law of conservation of energy, we get
1 2 1 2
mv0  mv  mgl (1  cos  )
2
2
Substituting the known values will give us
v0  7.6m / s
(c) The answer to (b) and (c) will decrease if
(Answer)
0 is increased by a few degrees.
9 (31) Figure 4-13 shows an 8.00kg stone at rest on a spring. The spring is
compressed 10.0cm by the stone. (a) What is the spring constant? (b) The stone is
pushed down an additional 30.0cm and released. What is the elastic potential
energy of the compressed spring just before that release? (c) What is the change in the
gravitational potential energy of the stone-Earth system when the stone moves from
the release point to its maximum height, measured from the release point?
Solution: (a) The key idea here is that we can use the Hooke’s
Law to find out the spring constant.
k
F mg 80 N


 800 N / m
x
x
0.1m
(Answer)
(b) When the stone is pushed down, the elastic potential energy
of the compressed spring can be solved as
Fig. 4-13 Problem 9
Ek 
1
1
k (x) 2   800 N / m  (0.4m) 2  64 J
2
2
(Answer)
(c) When the stone moves from the release point to its maximum height, according to
the work-kinetic energy theorem, the change in the gravitational potential energy of
the stone-Earth system will be
Wk  Wg 
1 2 1 2
mv  mv0  0
2
2
U  Ek  64 J
(Answer)
(d) Assuming the maximum height measured from the release point is h , we can get
h
U
64 J

 0.82m
mg 8.00kg  9.8m / s 2
(Answer)
10 (42) A stone with a weight of 5.29N is launched vertically from ground level
with an initial speed of 20.0m/ s , and the air drag in it is 0.265N throughout the
Chapter 4
Energy and Work
flight. What are (a) the maximum height reached by the stone and (b) its speed
just before it hits the ground ?
Solution: (a) There are two key ideas in this question. The first one is that the stone
can be seen as a particle and the free-body diagram of the stone as shown in Fig. 4-15.
We can assume the maximum height reached by the stone is hmax , so the work
Wg  mghmax , the work done by the air drag is W f   fhmax . The second key idea
is that when the stone reached the maximum height, the speed of the stone is zero, so
we can use the work-kinetic theorem as:
v
Wg  W f  E f  Ei
m
recasting this equation, we find
1
mghmax  fhmax  0  mvi2
2
Solving for hmax , and substituting known values, yields
hmax 
Fg
Fig. 4-14 Problem 10
Gvi2
5.29  202

 19.43m
2 g (G  f ) 2  9.8  (5.29  0.265)
(Answer)
(b) The key idea here is that when the stone is downward, the air drag on the stone is
upward, with the work-kinetic energy theorem, we can write as
Ghmax  fhmax 
1 2
mv f  0
2
Solving for v f and substituting known data, we will obtain
vf 
2hmax (G  f ) g
 19.02m
G
(Answer)
10 (45) In Fig.4-15, a block slides along a path that is without friction until the block
reaches the section of length L  0.75 m , which begins at height h  2.0 m on a
ramp of angle   30 . In that section , the coefficient of kinetic friction is 0.40 .
The block passes through point A with a speed of 8.0m/ s . If the block can reach

point B ( where the friction ends ), what is its speed there, and if it cannot , what is
its greatest height above A ?
Chapter 4
Energy and Work
Solution: (a) There are two key ideas. The first one is that we can assume the
block cannot reach point B ,
so the final speed is zero. The
second one is that the block and
the path can be seen as an
isolated
system and its
total energy can’t change. So
we can apply the law of
conservation of energy to this system:
4m
H
Fig. 4-14 Problem 10
Emec  Eth  0
The energy changes occurs between the initial state and the final state. The
change Emec is the sum of the change K in the kinetic energy and the change
U in the
gravitational potential energy of the system,
where.
1
K  0  mvA2
2
U  m g H
Eth   mg cos (2H  4)
Combining the equations above, we can write as
1 2
mv A  mgH   mg cos  (2 H  4)
2
Solving for H and substituting known data, we get
H
vA2
64

 2.7m  2m
2 g (1   cot  ) 2  9.8  (1  0.4 3)
So the block can reach the section with friction. Assuming the final speed at the lowest
part of the section is v f
1 2 1 2
mv f  mvA  mgh
2
2
During the section with friction, the acceleration of the block is
a   g sin    g cos 
Chapter 4
Energy and Work
Assuming the final speed of block on the section is zero, we will get
0  v2f  2as
s  1.3m  0.75m
So when the block reached the point B , its speed didn’t reduce to zero.
Solving it, we can obtain
vB2  v2f  2aL
Substituting known values, we can yield
vB  3.5m / s
(Answer)
11 (48) A 30 g bullet moving a horizontal velocity of 500m/ s comes to stop
12cm within a solid wall. (a) What is the change in the bullet’s mechanical energy ?
(b) What is the magnitude of the average force from the wall stopping it ?
Solution: (a) The key idea here is that when the bullet stops within a solid wall, the
speed of the bullet is zero. The bullet’s mechanical energy changes occur between the
initial state and the final state, the change
Emec  K 
1 2
mvi  0
2
Substituting known data gives us
Emec 
1 2 1
mvi   30 103  (500) 2  3750 J
2
2
(Answer)
(b) The key idea is that the work WF done on the bullet by the solid wall, the bullet
kinetic energy has an initial value of K 
1 2
mvi and a value of zero when the bullet
2
stops, so we can write the work-kinetic energy theorem for the bullet as
K f  Ki   Fs
Substituting known data, yields
F
Ki 3750 J

 31250 N
s
0.12m
(Answer)
12 (50) In Fig. 4-15, a chain is held in a frictionless table with one fourth of its length
hanging over the edge. If the chain has length L  28 cm and mass m  0.012kg ,
Chapter 4
Energy and Work
how much work is required to pull the hanging part back onto the table ?
Solution: The key idea here is that the chain has one fourth of its length hanging over
the edge. We can apply the work-kinetic energy theorem for the chain as
WF  WG  Ek
where the WF is the work done by the force F to pull the hanging part back onto the
table, the WG is the work done by one fourth of the chain’s gravity. The chain’s
initial kinetic energy and final kinetic energy are
zero. So the change of the kinetic energy
Ek  0
WF  m0 g
L
0
4
Substituting known values, we will obtain
Fig. 4-15 Problem 12
7
WF  m0 gL  0.003  9.8  102  1.03  103 J
2
(Answer)
13 (53) Resistance to the motion of an automobile consists of rode friction, which is
almost independent of speed, and air drag, which is proportional to speed–squared. For
a certain car with a weight of 12000N , the total resistant force F is given by
F  300  1.8v 2 , with F in newtons and v in meters per second. Calculate the
2
power (in horsepower) required to accelerate the car at 0.92 m / s when the speed
is 80km/ h .
Solution: There are three key ideas in this problem. First, we can relate the
acceleration a to the net force Fnet acting on the car with Newton’s second law, so
we can write as
Fnet  ma
Substituting known data, we get
Fnet 
G
12000
a
 0.92  1126.5 N
g
9.8
The second key idea is that the total resistance force
F
is given by
Chapter 4
Energy and Work
F  300  1.8v 2 , now we know the instantaneous speed is 22.22m / s , we
find
Fe  F  Fnet
where Fe is the force acting on the car. Substituting known data, we get
Fe  Fnet  F  1126.5  300  1.8  (22.22) 2  2315.4 N
The third key idea here is that the instantaneous power is equal F  v , so we can write
P  Fv cos
Substituting known values, we can yield
P  2315.4  22.22  51453W
14 (56) In Fig.4-16a, a block with a kinetic energy of 30 J is about to collide with a
spring at its relaxed length. As the block compress the spring, a frictional force
between the block and floor acts on the block. Figure 4-66bgives the kinetic energy
K ( x ) of the block and the potential energy U ( x ) of the spring as functions of
position x of the block, as the spring is compressed. What is the increase in thermal
energy of the block and the floor when (a) the block reaches position x  0.10m
and (b) the spring reaches its maximum compression ?
Fig. 4-16 Problem 14
Solution: (a) The key idea here is that the thermal energy of the block and the floor is
equal to the work done by the frictional force between the block and floor acting on
the block. So we need to find the work done by frictional force at x  0.10m in Fig.
Chapter 4
Energy and Work
4-66(a), at the position x  0.08m , the kinetic energy of the block is
K ( x)  22 J and the potential energy of the spring is U ( x)  2 J . Thus, at the
position x  0.08m , according to the law of conservation of
energy, we can obtain
K ( x)  U ( x)  W f  Ki  30 J
W f  30 J  22 J  2 J  6 J
Since W f   mgx   mg  0.08m , so f   mg 
Wf
x

6J
 75 N
0.08m
Then at x  0.10 , the thermal energy of the block and the floor is that
W   fx   mgx  75 N  0.10m  7.50 J
(b) The key idea here is when the spring reaches its maximum compression, the
kinetic energy of the block is zero and the spring has its maximum potential energy.
According to the law ot conservation of energy, we can get
Wf   U ( x)  30 J
in which,
W f   mgxmax ,U ( x)  
xmax
0
kxdx 
1 2
kxmax
2
At the position x  0.08m , we get
x
1
U ( x)   kxdx  kx 2  2 J
0
2
4J
4J
k 2 
 625 N / m
then
x
0.0064m2
1 2
1
2
30 J   mgxmax  kxmax
 (75 N ) xmax   (625 J / m 2 ) xmax
thus,
2
2
Simplifying it and solving for x , we will get
xmax  0.21m or xmax   0.45m
In Fig. 4-16(b), the direction of x is positive, so xmax  0.21m
15 (59) A 1.50kg snowball is shot upward at an angle of 34.0
（Answer）

to the horizontal
with an initial speed of 20.0m / s . (a) What is its initial kinetic energy? (B) By how
much does the gravitational potential energy of the snowball—Earth system change as
Chapter 4
Energy and Work
the snowball moves from launch point to the point of maximum height ? (c)
What is the maximum height ?
Solution: (a) According to the work-kinetic energy theorem, we can solve the kinetic
energy of the snowball as
Ek 
1 2 1
mv  1.50kg  (20.0m / s) 2  300 J
2
2
(Answer)
(b) The first key idea here is that as the snowball reaches the point of maximum height,
its kinetic energy is zero and gravitational potential energy will reach the maximum
value. So according to potential energy and the conservation of energy, we can write:
1 2
mv  U 0  U max  E
2
The second key idea here is that as the snowball at initial point, its gravitational
potential energy is zero, and the kinetic energy is
Ek 
1 2
mv  300 J
2
U  U max  300 J
so
(Answer)
(c) the maximum value of the potential energy is
U max  mghmax
Solving for hmax , and substituting known values, we get
hmax 
U max
300 J

 20.4m
mg 1.5kg  9.8m / s 2
(Answer)
16 (66) In Fig.4-17, a 1400kg block of granite is pulled up
an incline at a constant speed of 1.34m/ s by a cable and
winch. The indicated distance d1  40m and d 2  30m .
The coefficient of kinetic friction between the block and the
incline is 0.40. What is the power due to the force applied to
the block by the cable?
Fig. 4-17 Problem 16
Solution: A key idea here is that we can treat the granite as a particle. Another key
idea is that the particle moves with a constant speed. Thus. we can use equation
P  F  v to find the power due to the force applied to the block by the cable. From
Chapter 4
Energy and Work
the block’s free-body diagram in Fig. 4-17 a, we can find
F  f  mg sin   0
N  mg cos   0
f  N
d
tan   2
d1
N
F
f
G
Fig. 4-17a Problem 16
Solving for F , we get F 
mg
d12  d 22
(  d1  d 2 )
Substituting F with known values, we yield
P  F  v  Fv cos   Fv cos1800  Fv
P
1400kg  9.8m / s 2
(40m) 2  (30m) 2
(0.40  40m  30m) 1.34m / s
(Answer)
 1.69 104 J
17 (68)
A skier weighing 600 N goes over a
frictionless circular hill of radius R  20m (Fig.4-18).
Assume that the effects of air resistance on the skier are
negligible. As she comes up the hill, her speed is 8.0m / s
at point B, at angle   20 . (a) What is her speed at the
hilltop (point A) if she coasts without using her poles?
Fig. 4-18 Problem 17
(b) What minimum speed can she have at B and still coast to the hilltop? (c) Do the
answers to these questions increase, decrease, or remain the same if the skier weighs
700 N ?
Solutions: (a) The key idea is that the hill is frictionless and the effects of air
resistance on the skier are negligible, so the mechanical energy of the skier and hill
system is conservative. Choose point B as reference point. Now we have
1 2 1 2
mvB  mvA  mgR(1  cos  )
2
2
Substituting known values and solving for v A , we get
Chapter 4
Energy and Work
vA  vB2  2 gR(1  cos  )
 (8.0m / s) 2  2  9.8m / s 2  20m  (1  cos 20 )
(Answer)
 6.35m / s
(b) Assume the minimum speed can the skier have at A is v A . Obviously, vA  0 .
The minimum speed at B is
vB  2 gR(1  cos  )
 2  9.8m / s 2  20m  (1  cos 20 )
(Answer)
 4.86m / s
(c) We can see the speed at A and B has no relation with the mass of the skier, so
the answers remain the same if the skier weighs 700N .
18 (70) A 9.40kg projectile is fired vertically upward. Air drag decreases the
mechanical energy of the projectile—Earth system by 68.0KJ during the
projectile's ascent. How much higher would the projectile have gone were air drag
negligible?
Solution: The key idea here is that the total energy of the projectile is constant. During
the projectile ascent until it reaches its highest point, its kinetic energy transfers to its
potential energy and system’s thermal energy. If the air drag is negligible, there is no
thermal energy. Choose the tired point as the reference point, we have
1 2
mv  mgh  Eth
2
1 2
mv  mgh
2
Then we get
mg(h  h)  Eth
h  h 
Eth 68.0kJ

 7234m
mg
9.40kg
(Answer)
So the projectile would have gone 7234m higher were air drag negligible.
19 (73) A river descends 15m through rapids. The speed of the water is 3.2m/ s
upon entering the rapids and 13m / s upon leaving. What percentage of the
gravitational potential energy of the water—Earth system is transferred to kinetic
Chapter 4
Energy and Work
energy during the descent ? (Hint: Consider the descent of , say , 10kg of
water.)
Solution: The key idea here is that we can choose the point where the water leaves as
reference point. Then the change of the gravitational potential energy is
E p  mgh  10kg  9.8m / s 2 15m  1470 J
The change of the kinetic energy of the water is
E k 
1 2 1 2 1
mv2  mv1   10kg  [(13m / s) 2  (3.2m.s ) 2 ]  793.8 J
2
2
2
So the percentage of the potential energy transferred to kinetic energy is that
Ek
793.8 J
 100% 
 100%  54%
Ek
1470 J
(Answer)