Work and Energy

```WORK
In order for work to be done, three things are necessary:
•There must be an applied force.
•The force must act through a certain distance, called
the displacement.
•The force must have a component along the
displacement.
Work is a scalar quantity equal to the product of the
magnitudes of the displacement and the component of
the force in the direction of the displacement.
W=F.x
or
W = F cos  x
UNITS: N.m this unit is called a Joule (J)
As long as this person does
not lift or lower the bag of
groceries, he is doing no work
on it. The force he exerts has
no component in the direction
of motion.
Work done by forces that
oppose the direction of motion,
such as friction, will be negative.
Centripetal forces do no work, as they are always
perpendicular to the direction of motion.
If the force acting on an object varies in magnitude
and/or direction during the object’s displacement,
graphical analysis can be used to determine the work
done. The Force is plotted on the y-axis and the distance
through which the object moves is plotted on the x-axis.
The work done is represented by the area under the
curve.
5.1 A push of 200 N moves a 100 N block up a 30inclined plane.
The coefficient of kinetic friction is 0.25 and the length of the plane
is 12 m.
a. Find the work done by each force acting on the block.
W
FA = 200 N
FG = 100 N
θ = 30˚
μ = 0.25
x = 12 m
Forces acting: Ff FA FG and FN
FN does NO work.
Ff = μ F N
= μ FG cos 30˚
= 0.25 (100) cos 30˚ = 21.6 N
W FA = FA x
= 200 (12)
= 2400 J
FN
FA
θ
FGy
Ff
FG
FGx
WFf = - Ff x
= - 21.6 (12)
= - 259.2 J
WFG = FG x = -FGx x = - FG sin30˚x
= - 100 sin 30˚ (12)
= - 600 J
b. Show that the net work done by these forces is the same as the
work of the resultant force.
Net work:
ΣW = 2400 - 259.2 - 600
= 1540.8 J
FN
FA
θ
FGy
Ff
FG
FGx
The resultant force:
ΣFx = FA - Ff - FGx
= 200 - 21.6 - 50
= 128.4 N
WF = F x . x
= 128.4 (12)
= 1540.8 J
ENERGY
Energy is that which can be converted into work. When
something has energy, it is able to perform work or, in a
general sense, to change some aspect of the physical
world.
In mechanics we are concerned with two kinds of
energy:
KINETIC ENERGY: K, energy possessed by a body by
virtue of its motion.
1
K  mv
2
Units: Joules (J)
2
POTENTIAL ENERGY: PE, energy possessed by a
system by virtue of position or condition.
PE = m g h
Units: Joules (J)
WORK-ENERGY PRINCIPLE:
The work of a resultant external force on a body is equal
to the change in kinetic energy of the body.
W=K
Units: Joules (J)
W = PE
5.2 What average force F is necessary to stop a 16 g bullet
traveling at 260 m/s as it penetrates into wood at a distance of
12 cm?
WE
1
2
2
W = ΔK 
vf = 0 m/s
m = 0.016 kg
vo = 260 m/s
x = 0.12 m
2
m(v f  v o )
1 2
Fx   mvo
2
2
o
2
mv
(0.016)(260)
F 

= - 4506.7 N
2x
2(0.12)
CONSERVATIVE AND NON-CONSERVATIVE FORCES
The work done by a conservative force depends only on
the initial and final position of the object acted upon. An
example of a conservative force is gravity.
The work done equals
the change in potential
energy and depends
only on the initial and
final positions above
the ground and NOT on
the path taken.
Friction is a non-conservative force and the work done
in moving an object against a non-conservative force
depends on the path. For example, the work done in
sliding a box of books against friction from one end of a
room to the other depends on the path taken.
For mechanical systems involving conservative forces,
the total mechanical energy equals the sum of the
kinetic and potential energies of the objects that make
up the system and is always conserved.
PE  K
A roller-coaster car moving without friction illustrates the
conservation of mechanical energy.
In real life applications, some of the mechanical energy
is lost due to friction. The work due to non-conservative
forces is given by:
WNC = ΔPE + ΔK
or
WNC = Ef - Eo
5.3 A ballistic pendulum apparatus has a 40-g ball that is caught by
a 500-g suspended mass. After impact, the two masses rise a
vertical distance of 45 mm. Find the velocity of the combined
masses just after impact.
COE
m1 = 0.04 kg
m2 = 0.500 kg
h = 0.045 m
K0 = PEf
m1 = 0.04 kg
m2 = 0.500 kg
h = 0.045 m
K0 = PEf
PEf = (m1+m2) ghf
= (0.04+0.500)(9.8)(0.045)
= 0.24 J
K0 = PEf = 0.24 J
v0 
1
K 0  mT v02
2
2(0.24)
2Ko

= 0.94 m/s
0.54
mT
5.4 The tallest and fastest roller coaster in the world is the Steel
Dragon in Japan. The ride includes a vertical drop of 93.5 m. The
coaster has a speed of 3 m/s at the top of the drop.
a. Neglect friction and find the speed of the riders at the bottom.?
A
COE
vA = 3 m/s
hA = 93.5 m
hB = 0 m
B
At point A: PEA + KA
At point B: KB
PEA + KA= KB
1
1
2
2
mgh  mv A  mv B
2
2
v B  v A2  2 gh  (3) 2  2(9.8)(93.5) = 42.9 m/s
b. Find the work done by non-conservative forces on a 55 kg
rider during the descent if the actual velocity at the bottom is
41 m/s.
v = 3 m/s
A
WNC = Ef - E0
= KB - (PEA + KA)
vB = 41 m/s
hA = 93.5 m
hB = 0 m
m = 55 kg
1 2 
1 2
 mvB   mghA  mvA 
2
2


1
1

2
2
 (55)(41)   (55)9.8(93.5)  (55)(3) 
2
2


= - 4416.5 J
5.5 A 20-kg sled rests at the top of a 30˚ slope 80 m in length.
If μk= 0.2, what is the velocity at the bottom of the incline?
m = 20 kg
θ = 30°
r = 80 m
μk= 0.2
WNC = Ef - Eo
= Kf - PE0
COE
m = 20 kg
θ = 30°
x = 80 m
μk= 0.2
x
h
h = x sin θ
h = 80 sin 30°
= 40 m
Ff = μkFN = μk Fgy
= μk Fgcos30°
= (0.2)(20)(9.8)cos30°
= 34 N
WNC= Kf - PE0
Kf = PE0 + WNC
= 7840 - 2720 = 5120 J
PE0 = mgh0
= 20(9.8)(40)
= 7840 J
WNC = - Ff r
= - 34 (80)
= - 2720 J
1 2
K f  mv f
2
2K
2(5120)
vf 
= 22.6 m/s

m
20
POWER
Is the rate at which work is performed.
W Fr
P

 Fv
t
t
J
UNITS:
=W
s
P = work/time
= Watt
The difference between walking and
running up these stairs is power.
The change in gravitational potential
energy is the same.
5.6 A 1100-kg car starting from rest, accelerates for 5.0 s. The
magnitude of the acceleration is 4.6 m/s2. What power must the
motor produce to cause this acceleration?
m = 1100 kg
vo = 0 m/s
t=5s
a = 4.6 m/s2
vf = vo + at
= 0 + 4.6 (5)
= 23 m/s
F = ma
= (1100)(4.6)
= 5060 N
The average velocity is: 23/2 = 11.5 m/s
P = Fv
= (5060)(11.5)
= 5.82x104 W
ELASTIC FORCE
The force Fs applied to a spring to stretch it or to
compress it an amount x is directly proportional to x.
Fs = - k x
Units: Newtons (N)
Where k is a constant called the spring constant and is
a measure of the stiffness of the particular spring. The
spring itself exerts a force in the opposite direction:
This force is sometimes called restoring force because
the spring exerts its force in the direction opposite to
the displacement. This equation is known as the spring
equation or Hooke’s Law.
1
A  bh
2
1
 ( x)(kx)
2
1 2
 kx
2
The elastic potential energy is given by:
PEs = ½ kx2 Units: Joules (J)
5.7 A dart of mass 0.100 kg is pressed against the spring of a toy
dart gun. The spring (k = 250 N/m) is compressed 6.0 cm and
released. If the dart detaches from the spring when the spring
reaches its normal length, what speed does the dart acquire?
m = 0.1 kg
k = 250 N/m
x = 0.06 m
PEs = K
½ kx2 = ½ mv2
v
2
kx
250(0.06) 2

0.1
m
= 3 m/s
page 108
Problem 4
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