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Sub-Topics
• Work and standard units
•Power concept and Power calculation
•Kinetic energy concept
• Potential energy
•The law of conservation of energy
Work
• an activity involving a force and movement in the
direction of the force or when a force acts upon an
object to cause a displacement of the object
• Work Done by a constant force: product of the
magnitudes of the displacement and the component
of the force parallel to that displacement.
W = F|| d = (F cos θ )d
F = force (N)
d = displacement (m)
W = work (Nm @ J)
θ: angle between displacement and force
vectors
(a) If there is no displacement, no work is done: W = 0.
(b) For a constant force in the same direction as the displacement, W = Fd.
(c) For a constant force at an angle to the displacement, W = (F cos θ)d.
EXAMPLE 1
A students holds a book, which
has a mass of 1.5 kg, out a
second-story dormitory until her
arm is tired, then she releases it
(a)How much work is done on the
book by the student in simply
holding it out the window?
(b)How much work is done by the
force of gravity during the time in
which the book falls 3.0 m?
Solution
Given
vo = 0 (initially at rest)
m = 1.5 kg
d= 3.0 m
(a)W = 0
(She does no work on the book, even though she exerts an
upward force on the book, but the displacement is zero)
(b) W = F . d = (m. g)d = (1.5kg)(9.8m/s2)(3.0m) = 44 J
(While the book is falling, the force acting is gravity, which is
equal in magnitude to the weight of the book)
Work, Force & Displacement
Total or Net work
• Work done by all the forces acting on the object or scalar
sum of all those quantities of work
EXAMPLE 2
A 0.75kg block slides with a uniform velocity down a 20o inclined plane
(refer Figure below)
(a)How much work is done by force of friction on the block as it slides the
total length of the plane
(b) What is the net work on the block?
Solution
Given: m = 0.75kg
θ = 20o
(a)
W f = f k (cos180o )d = − f k d
L = 1.2 m
The angle 180o indicates the force and displacement are in opposite
directions
Distance:
d=
L
cos θ
f k = mg sin θ
Friction force:
So, work done by frictional force:
 L 
W f = − (mg sin θ )
 = − mgL tan θ
 cos θ 
W f = − (0.75kg )(9.8m / s 2 )(1.2m)(tan 20o ) = − 3.2 J
Solution
(b) To find net work, we need to calculate the work done to the
object by all forces
L
Distance:
d=
cos θ
Gravity force:
f g = mg sin θ
So, work done by gravity force:
 L 
Wg = (mg sin θ )
 = mgL tan θ = + 3.2 J
 cos θ 
Net Work:
Wnet = Wg + W f = + 3.2 J + − 3.2 J = 0
Work Done by a Variable Force
• Forces in practice will often vary, they change in magnitude or
angle with time or position. E.g. Someone might push harder
and harder on an object to overcome the force of static
friction, until the applied force exceeds fsmax
• Example of variable force is the force applied to the spring
(refer fig 5)
Fs = − k∆ x = − k(x − xo )
• For ideal spring force:
• Thus,
W=
Fs = − kx
1
2
kx
2
where
k = force constant
x= distance
Figure 5
(a) An applied force Fa stretches the spring, and the spring exerts an equal and
opposite force Fs on the hand.
(b) The magnitude of the force depends on the change Δx in the spring’s length.
This change is measured from to the end of the unstretched spring at x0.
Note: In Fig. 5, the hand applies a variable force Fa in stretching the spring. At
the same time, the spring exerts an equal and opposite force Fs on the hand.
EXAMPLE 3
A 0.15kg mass is attached
to a vertical spring and
hangs at rest a distance of
4.6cm below its original
position. An additional
0.5kg mass is then
suspended form the first
mass and allowed to
descend to a new
equilibrium position. What
is the total extension of
the spring?
Solution
Given:
m1 = 0.15kg
m2 = 0.5kg
For ideal spring force:
x1 = 4.6cm = 0.046m
x=?
Fs1 = kx1 = m1 g
Solving to find spring constant, k:
m1 g 0.15kg (9.8m / s 2 )
k=
=
= 32 N / m
x1
0.046m
Thus,
Fs = (m1 + m2 ) g = kx
(m1 + m2) g (0.15kg + 0.5kg )(9.8m / s 2)
x=
=
= 0 .2 m
k
32 N / m
Energy
• the ability to do work. When work is done,
energy is transferred
• forms of energy e.g. electrical, chemical heat,
nuclear, mechanical etc
• Units: Joules (J)
• Types of energy
– Kinetic
– Potential
• Energy comes in many forms and always
conserved.
Forms of Energy
Radiant
Thermal
Electrical
Nuclear
Chemical
Sound
Mechanical
Mechanical
Magnetic
Kinetic energy concept
• Kinetic energy is the ability to do work through motion. Also
called as energy of motion
Remember
2
2
2
2
v f = vo + 2 a ( ∆ x )
vf
v
− o = a∆ x
2
2
Multiply both side with mass;
1
1
2
2
mv f − mvo = ma∆ x
2
2
Thus, kinetic energy;
K=
1 2
mv
2
K = kinetic energy
m= mass
v = velocity
vf = final velocity
vo = initial velocity
The work done on a block by a constant force in moving it along a horizontal
frictionless surface is equal to the change in the block’s kinetic energy: W = ΔK.
EXAMPLE 4
What is the kinetic energy of a 4 kg shot-put thrown by an
athlete at a speed of 15 m/s?
Solution
Kinetic energy = 1/2 mv2
= 1/2 x 4 kg x (15 m/s)2 = 1/2 x 4 kg x 225 m2/s2
= 450 J
Potential Energy
• Potential energy exists whenever an object which
has mass has a position within a force field
(gravitational, magnetic, electrical).
• An object having potential energy has potential to
do work
• Often called the energy of position
• We will focus primarily on gravitational potential
energy (energy an object has because of its height
above the Earth)
gravitational potential energy
Gravitational potential energy
U = mgy
m = mass
g = acceleration due to
gravity
y =distance
Kinetic & Potential Energy Relationship
If force depends on distance,
∆P E = −F ∆x
For gravity (near Earth’s surface)
∆P E = m gh
EXAMPLE 5
What is the potential energy of a 12 kg mass raised to a
height of 25 m?
Solution
Potential Energy = weight x height change
Weight (m x g) = 12 kg x 10 N/kg = 120 N
Height change, y = height at end - height at start = 25 - 0 = 25 m
Potential energy (U) = m x g x y = 120 N x 25 m = 3000 J
EXAMPLE 6
A diver of 75 kg drops from a board
10.0 m above the water surface, as
in the Figure. Find his speed 5.00 m
above the water surface. Neglect
air resistance.
9.9 m/s
Potential Energy
The law of conservation of energy
• The Law of Conservation of Energy States:
the total energy of an isolated system is always
conserved
• Energy is neither created nor destroyed; it is
converted from one form to another
• example in a nuclear power station. – Nuclear energy is converted into heat.
– Heat boils water to steam
– Heat in the steam is converted into kinetic
energy in the turbines
– Which is converted into electrical energy in the
generator
Conservative & Nonconservative Force
Conservative
Nonconservative
Work done by it in moving an
object is independent of the
object’s path
Work done by it in moving an
object does depend on the
object’s path
depends only on the initial and
final positions of an object
E.g: Friction, a longer path would
produce more work done by
friction than a shorter one, more
energy would be lost
Total energy is constant
E = Eo, K + U = Ko - Uo
Total energy is not constant
E = Eo
Allows to conserve or store
energy as potential energy
EXAMPLE 8
A skier with a mass of 80kg starts from rest at the top of a slope
and skis down from an elevation of 110m. The speed of the
skier at the bottom of the slope is 20m/s
(a)Show that the system is nonconservative
(b)How much work is done by the nonconservative force of
friction
Solution
Given;
m = 80kg
vo = 0
v = 20m/s
yo = 110m
(a) If the system is conservative, total mechanical energy is constant
Energy at the top
Eo = U = mgyo = (80kg)(9.8m/s2)(110m) = 8.6 x104 J
Energy at the bottom
1
1
E = K = mv 2 = (80kg )(20m / s ) 2 = 1.6 x10 4 J
2
2
E ≠ Eo
, so the system is nonconservative.
(b) The amount of work done by the nonconservative force of friction is
equal to the change in the mechanical energy, or amount of
mechanical energy lost
Wnc = E − Eo = (1.6 x10 4 J ) − (8.6 x10 4 J ) = − 7.0 x10 4 J
Power Concept
•
•
•
•
the rate of doing work or the rate of using energy
Power = energy transferred (J) = work done (J) time taken (s)
time taken (s)
SI Unit : 1 W (watt) = 1 Js-1 (joule per second)
British Unit: 1 hp (Horsepower) = 746 W
• Work done = force x distance moved: W = Fs
• Power = energy ÷ time: P = W/t
• Speed = distance ÷ time: v = s/t
• So we can write:
P = W/t = Fs/t
• Therefore:
Power (W) = force (N) × speed (m/s)
EXAMPLE 9
A crane hoist lifts a load of 1000kg a vertical distance of
25m in 9 s at a constant velocity. How much useful work is
done by the hoist each second?
Solution
Given: m = 1000kg
d = 25m
P=
t = 9.0 s
W Fd mgd
=
=
t
t
t
(1000kg )(9.8m / s 2 )(25m)
=
= 27 x10 4 W = 27 kW
9.0 s
EXAMPLE 10
The motors of two vacuum have net power outputs of 1.00
hp and 0.50 hp, respectively.
(a)How much work in joules can each motor do in 3.0 min?
(b)How long does each motor take to do 97.0 kJ of work
Solution
Given; P1= 1.00 hp = 746 W
P2 = 0.50 hp = 373 W
t = 3 min = 180 s
W = 97.0 kJ
(a) Since P = W/t
W1 = P1t = (746 W)(180s) = 1.34 x 105 J
and
W2 = P2t = (373W)(180s) = 0.67 x 105 J
(b) Times, t = W/P
t1 = W/P1 = (97 x 103 J)/(746W) = 130 s
and
t2 = W/P2 = (97x 103 J)/(373 W) = 260 s
 Note that, smaller motor takes more time and does
less work than larger motor
Efficiency
Relates work output to energy (work) input as
a percent
ε = Work output ( x100% ) = Wout ( x100% )
Energy input
Ein
ε = Power output ( x100% ) = Pout ( x100% )
Pin
Power input
EXAMPLE 11
The motor of an electric drill with an efficiency of 80%
has a power input of 600kW. How much useful work is
done by the drill in 30s?
Solution
Given; ε = 80 % Pin = 600kW
t = 30s
Pout = εPin = (0.8)(600kW) = 480 kW
From, P = W/t
wout = Poutt = (480 kW)(30 s) = 14.4 x 106 J