A. Z. ALZAHRANI Chapter 8: Conservation of Energy
This chapter actually completes the argument established in the previous chapter and
outlines the standing concepts of energy and conservative rules of total energy. I will
divide the chapter into two main parts: conservative and nonconservative forces and
conservation of energy.
A- Conservative and Nonconservative Forces
There are two basic properties of the conservative forces that are related to the work
done by this force. These are:
• If the work done by the force is independent of the path, the force is said to
be a conservative force. Otherwise, it is nonconservative.
• The work done by a conservative force in any closed path is zero. The closed
path means a zero displacement.
As examples of conservative forces, the gravitational and spring forces do. On the
other side, friction force gives an example of nonconservative forces.
B- Conservation Law of Mechanical Energy
The law of conservation of energy states that: the total mechanical energy of a
system remains unchanged (constant) in any isolated system of objects that
interact only through conservative forces. In other words, it states that: the energy
may neither be created nor destroyed, but transferred from a form of energy to
another.
The total mechanical energy, E, is defined as:
E=K+U
Where K is the total kinetic energy and U is the total potential energy of the system.
A. Z. ALZAHRANI The conservation law of energy is , therefore,
Ei = Ef
Where Ei and Ef are the initial and final energies, respectively. That implies to
Ki + Ui = Kf + Uf
The potential energy is defined (from chapter 7) as:
U= Wg = F. d = mg . d=mg d cosθ
Whereas the kinetic energy is defined as:
K = ½ m v2
In the following diagram, kinetic and potential energies are calculated for a car
starting its motion from rest at the top of a cliff. The total energy of the car is
conserved at each point till it reaches the bottom of the cliff.
A. Z. ALZAHRANI Examples:
1. Using the work-energy theorem, show that W= -∆U.
Solution
The work-energy theorem states that
W = ∆K
However, the law of conservation of energy is:
Ki + Ui = Kf + Uf
It can be rewritten as
Kf - Ki = Ui - Uf = -(Uf – Ui)
Or
∆K = - ∆U
Therefore,
W = - ∆U
2. Find the height from which you would have to drop a ball so that it
would have a speed of 9.0 m/s just before it hits the ground.
Solution
The initial energy of the system is
Ei = m gh
The final energy of the system is
Ef = ½ mv2
Therefore,
½ mv2= m gh
A. Z. ALZAHRANI Hence,
h= v2/2g = 9.02/19.6 = 4.1 m
3. A ball is thrown vertically upwards and reaches a maximum height of
19.6 m. Calculate its initial speed.
Solution
The initial energy of the system is
Ei = ½ mv2
The final energy of the system is
Ef = m gh
Therefore,
½ mv2= m gh
v = [2 gh]0.5=[2 × 9.8 × 19.6]0.5=19.6 m/s
3. A block having a mass of 4.0 kg is given an initial velocity vA =2.0 m/s
to the right and collides with a spring of force constant k= 64.0 N/m.
Calculate the maximum compression of the spring after the collision,
assuming the surface is frictionless.
Solution
8.5 Work Done by Nonc
x=0
vA
(a)
The initial energy, Ei, is
(b)
E = –12 mvA2
!
vB
Ei "
= Ki + Ui
xB
E = –12 mv B2 + –12 kx B2
A. Z. ALZAHRANI There is only kinetic energy, therefore
P
Ei = ½ mv
2
The final energy, Ef, is
12. A mass m starts from rest and slides a distance d down a
frictionless incline ofEangle
!. While sliding, it contacts
f = Kf + Uf
an unstressed spring of negligible mass, as shown in FigWhile we have
askedThe
about
maximum
compression,
that
ure P8.11.
massthe
slides
an additional
distance x as
it means the
is brought momentarily to rest by compression of the
body will momentarily
stop. Therefore, at this point, there is only
spring (of force constant k). Find the initial separation
d between
and the
spring.
potential energy
due tothe
themass
spring.
Hence,
h
Ef = ½ kx2
m = 3.00 kg
From the conservation law, we have
Ei = Ef
d
Therefore,
k = 400 N/m
½ mv2
= ½ kx2
cal spring
downward
θ = 30.0°
ter the blo
x = (mv2/k)0.5 =(4 × 22/64)0.5 =0.5 m
leaves the
point of re
Problems
11
and
12.
Figure
P8.11
4. A 4.00 kg particle is freely released from point (A) and slides on18.
theDave John
decathlon
frictionless. Determine (a) the particle’s speed at points (B) and (C) and
jump with
13. A particle of mass m " 5.00 kg is released from point !
(b) the total and
work
done
force of track
gravity
in moving
slides
onby
thethe
frictionless
shown
in Figurethe particle fromHow far u
the jump?
P8.13. Determine (a) the particle’s speed at points "
(A) to (C). and # and (b) the net work done by the force of gravity
19. A 0.400-kg
reaches a
in moving the particle from ! to #.
Solution
position a
energy me
mechanic
ergy to the
! m
when the
20. In the dan
"
student ju
That gives
#
5.00 m
3.20 m
2.00 m
Figure P8.13
A. Z. ALZAHRANI (a) The initial energy at point (A), EiA, is
EiA = KiA + UiA
Because the body starts from rest, there is only potential energy
Ei = UiA
The final energy at point (B), EfB, is
EfB = KfB + UfB
At this point we have both kinetic and potential energies. From the
conservation law, we find
EiA = EfB
Or
UiA = KfB + UfB
That means
KfB = UiA - UfB
Therefore,
½ mvB2 = mg (dA -dB)
vB = [2 g (dA -dB)]0.5= [2×9.8×(5.0-3.2)]0.5 =5.95 m/s
The final energy at point (C), EfC, is
EfC = KfC + UfC
However,
EiA = EfC
Or
UiA = KfC + UfC
That means
KfC = UiA - UfC
A. Z. ALZAHRANI 242
CHAPTER 8
Therefore,
2
(b) The
Potential Energy and Conservation of Ene
½ mv
= ankles,
mg (dAas–d
elastic cord attached
toChis
shown
in Figure
C)
P8.20. The unstretched
length of the cord is 0.5
25.0 m, the
0.5
vCstudent
= [2 gweighs
(dA –d700
)]
=
[2×9.8×(5.0-2.0)]
C N, and the balloon is 36.0 m=7.68
above m/s
the surface of a river below. Assuming that Hooke’s law
workdescribes
done bythe
thecord,
gravity
between points A and C is
calculate the required force constant
if the studentW
is to =
stop
safely 4.00
m above
- ∆U
= - (U
– U the
) river.
AC
AC
C
A
cal circular a
mass m and h
form, startin
with respect
former’s bod
she does not
air resistance
make an ang
former must
21. Two masses are connected by a light string passing over a
light frictionless pulley, as shown in Figure P8.21. The
5.00-kg
mass is released from rest. Using the law of conW
AC = - mg (dC – dA)= - 2×9.8×(5.0 -2.0) = -58.8 J
servation of energy, (a) determine the speed of the 3.00F
kg mass just as the 5.00-kg mass hits the ground and (b)
in order to h
find the maximum height to which the 3.00-kg mass rises.
22. Two are
masses
are connected
a light
stringwhich
passing is
over
5. Two masses
connected
by a bylight
string,
passing overthe
a force req
is twice the p
a light frictionless pulley, as shown in Figure P8.21. The
light frictionless
The
mass m
from rest. Using the law
1 ismreleased
mass pulley.
m1 (which
is greater
than
2) is released from rest.
Using the law of conservation of energy, (a) determine
of conservation
of energy, determine the speed of m2 just as m1 hits the
the speed of m 2 just as m1 hits the ground in terms
of
m 1, m 2, and h, and (b) find the maximum height to
ground.
which m 2 rises.
Therefore
Solution
m1 " 5.00 kg
m2 " 3.00 kg
h " 4.00 m
Problems
21 andof
22.a similar acceleration.
Figure under
P8.21 the
The two masses will move
influence
26. After its rele
coaster car m
Therefore their velocities will be the same when the mass m1 hits the
roller coaste
23. A 20.0-kg cannon ball is fired from a cannon with a
ground. Hence
analysing
each
of
them,
separately,
gives
the
followings:
of radius 20.
muzzle speed of 1 000 m/s at an angle of 37.0° with the
loop: At the
horizontal. A second ball is fired at an angle of 90.0°.
and feel weig
Use the law of conservation of mechanical energy to
coaster car a
find (a) the maximum height reached by each ball and
speed of the
(b) the total mechanical energy at the maximum height
(c) at positio
for each ball. Let y " 0 at the cannon.
A. Z. ALZAHRANI For the mass, m1, the initial kinetic and potential energies are
Ki1 = 0
And
Ui1 = -m1 gh
Therefore,
Ei1 = -m1 gh
The final kinetic and potential energies (when it hits the ground) are
Kf1 = ½ m1v2
And
Uf1 = 0
Therefore,
Ef1 = ½ m2v2
For the mass, m2, the initial kinetic and potential energies are
Ki2 = Ui2 = 0
Therefore,
Ei2 = 0
The final kinetic and potential energies are
Kf2 = ½ m2v2
And
Uf2 = - m2 gh
Therefore,
Ef2 = ½ m2v2 - m2 gh
The initial total energy of the system is
Ei = Ei1 + Ei2 = -m1 gh
A. Z. ALZAHRANI The final total energy of the system is
Ef = Ef1 + Ef2 = ½ m1v2 + ½ m2v2 - m2 gh
It is known from the conservation law of energy that
Ei = Ef
That implies to
½ m1v2 + ½ m2v2 - m2 gh = -m1 gh
Therefore,
½ (m1+ m2) v2= (m2 – m1) gh
v = [2 (m2 - m2)/ (m1 + m2) gh]0.5
That gives
v = [2×9.8×4.0×(5.0-3.0)/(5.0+3.0)]0.5 =4.43 m/s
6. Two bodies are connected by a cord, which passes through a small
pulley. The coefficient of friction between the 3.00-kg block and the
surface is 0.4. Estimate the speed of the 5.00-kg ball when it has fallen
1.50 m?
Problems
243
Solution
3.00 kg
4
5.00 kg
Figure P8.31
For the mass, m1=5.00 kg, the initial energy
Ui1and
= 0coasts down from the
i1 =rest
32. A 2 000-kg car startsK
from
top of a 5.00-m-long driveway that is sloped at an angle
of 20.0° with the horizontal. If an average friction force
of 4 000 N impedes the motion of the car, find the
A. Z. ALZAHRANI Therefore,
Ei1 = 0
The final kinetic and potential energies are
Kf1 = ½ m1v2
Uf1 = -m1 gh
Therefore,
Ef1 = ½ m1v2 - m1 gh
For the mass, m2=3.00 kg, we know that the total work is equivalent to
the change in the kinetic energy
W = ∆K = - ∆U
The work done by the friction force only, therefore
W = -µkm2 gd
Hence
∆K= -µkm2 gd
From the conservation law of energy we get
½ m1v2 - m1 gh = -µk m2 gd
Or
v = [2 (m1 - µkm2)/ (m1 + m2) gd]0.5
v = [2 (5.0 – 0.4 × 3.0)/ (5.0+3.0) × 9.8 ×1.5]0.5 =3.73 m/s
A. Z. ALZAHRANI 7. A motorcyclist is trying to jump across the valley by driving
horizontally off the cliff at a speed of 38.0 m/s. Find the speed with which
the cycle strikes the ground on the other side, as shown in the figure.
Solution
The initial energy at the top, Ei, is
Ei = Ki + Ui = ½ mv02 - mg h0
The final energy at the bottom, Ef, is
Ef = Kf + Uf = ½ mvf2 - mg hf
From the conservation law, we find
½ mv02 - mg h0 = ½ mvf2 - mg hf
Then we get
vf = [v02 – 2 g (h0 - hf)]0.5
vf = [38.02 – 2 × 9.8 (70.0 – 35.0)]0.5 =27.53 m/s
A. Z. ALZAHRANI 8. A person is sitting on a sledge at the top of a 23.7 m tall hill. Determine
their speed when they reach the bottom of the hill.
Solution
The initial energy at the top, Ei, is
Ei = Ki + Ui = 0 + mg h
The final energy at the bottom, Ef, is
Ef = Kf + Uf = ½ mv2 + 0
From the conservation law, we find
½ mv2 = mg h
Then we get
v = [2 g h]0.5
v = [2 × 9.8 × 23.7]0.5 = 21.6 m/s