Work and Energy
Definitions
• Work
– the product of force and the component of
displacement in the direction of the force.
- work is a scalar quantity
- Without motion there is no work.
W=F∙d
The unit of work is the newton meter, which is called a
joule (J)
(in honor of English Scientist James Prescott Joule)
Example
• How much work is done on an object if a force
of 30 Newtons [south] displaces the object
200 meters [south]?
• Solution:
W = F∙d
= (30 N[S])(200 m [S])
= 6000 J
Work (cont’d)
• Suppose force and displacement are not in the
same direction.
• Work is defined to be the product of the force
in the direction of the displacement and the
displacement
F
W = (Fcos θ)∙d
object
θ
d
example
• As Alex pulls his red wagon down the
sidewalk, the handle of the wagon makes an
angle of 60 degrees with the pavement. If Alex
exerts a force of 100 Newtons along the
direction of the handle, how much work is
done when the displacement of the wagon is
20 meters along the ground?
Power
Definition
• Power – the rate at which work is done
W Fd
P

 Fv
t
t
Power is also a scalar quantity, and its unit is
Joules per second (J/s), also known as a
watt(W).
Example
• If 300 J of work is performed on an object in
1.0 minute, what is the power expended on
the object?
P=W/t
P = 300J / 60 sec
P=5W
Example 2
• A 200 N force is applied to an object that
moves in the direction of the force. If the
object travels with a constant velocity of
10m/s, calculate the power expended on the
object.
W Fd
P

 Fv
t
t
P = 200N (10 m/s)
P = 2000 W
Hw
• Read pg. 80-83 Do # 1-25
Energy
Mechanical Energy
Kinetic Energy
• Work done on an object changes its Kinetic
Energy.
– Therefore,
W = KEf – KEi = ΔKE.
• The formula for Kinetic Energy is
KE = ½ mv2
Example
• A 10 kg object subjected to a 20. N force
moves across a horizontal, frictionless surface
in the direction of the force. Before the force
was applied, the speed of the object was
2.0m/s. When the force is removed the object
is traveling at 6.0 m/s. Calculate the following
quantities: (a) KEi , (b) KEf , (c)ΔKE, (d) W, and
(e) d.
Solution
(a)
KEi
= ½ mvi2
= ½ (10. kg)(2.0 m/s)2
= 20. J
(b)
KEf
= ½ mvf2
= ½ (10. kg)(6.0 m/s)2
= 180 J
Solution (cont’d)
(c)
ΔKE = KEf – KEi
= 180 J – 20. J
= 160 J
(e)
W = F∙d
d = W/F
= 160 J / 20. N
= 8.0 m
(d)
W = ΔKE
= 160 J
Gravitational Potential Energy
• An object decreases its gravitational Potential
Energy (PE) as it moves closer to the Earth
• To calculate the change in PE of an object we
measure the work done on the object. The
force needed to overcome gravity is Fg = mg.
Therefore, since W = Fg ∙d , PE is defined as
ΔPE = mg Δ h
where Δh represents change in vertical
displacement above the earth.
Example
• A 2.00 kg mass is lifted to a height of 10.0 m
above the surface of the Earth. Calculate the
change in the PE of the object.
• Solution
ΔPE = mg Δ h
= (2.00 kg) (9.8 m/s2)(10.0m)
= 196 J
NOTE
• For a change in gravitational energy to occur,
there must be a change in the vertical
displacement of an object; if it is moved only
horizontally, the ΔPE = 0.
• If an object is moved up an inclined plane, its
potential energy change is measured by
calculating only its vertical displacement; the
horizontal part does not change its PE
Home work
• Review book : read pg 84 Do # 26-38 (PE)
– Read pg 92-93 Do #57 -66 (KE)
Conservation of Mechanical
Energy
Conservation of Mechanical Energy
• In a system, the sum of PE and KE (the total
mechanical energy) is constant (i.e.
conserved); a change in one is accompanied
by an opposite change in the other.
ΔPE = -ΔKE
PEi + KEi = PEf +KEf
example
• A 0.50 kg ball is projected vertically and rises
to a height of 2.0 meters above the ground.
Calculate:
(a) the increase in the ball’s PE
(b) the decrease in the ball’s KE
(c) the initial KE
(d) the initial speed of the ball
solution
(a) The increase in the ball’s PE
ΔPE = mg Δ h
= (0.50 kg)(9.8 m/s2)(2.0 m)
= 9.8 J
(b) The decrease in the ball’s KE
ΔKE = -ΔPE
= -9.8 J
Solution cont’d
(c) The initial KE
- Recall that at its highest point, the speed of
the ball is zero; therefore its KE is zero. So the
initial KE represents the change in KE of the
object .
ΔKE = KEf – KEi
-9.8J = 0 - KEi
KEi = 9.8 J
Sol’n cont’d
(d) The initial speed of the ball
KEi = ½ mvi2
Solving for vi = sqrt (2KEi /m)
= sqrt [ 2(9.8J)/(0.50kg) ]
= 6.3 m/s
Pendulum
Observe that the falling motion of the bob is
accompanied by an increase in speed. As the bob
loses height and PE, it gains speed and KE; yet the
total of the two forms of mechanical energy is
conserved
example
• A pendulum whose bob weighs 12 N is lifted a
vertical height of 0.40 m from its equilibrium
position. Calculate:
(a) change in PE between max height and
equilibrium height
(b) Gain in KE and
(c) the velocity at the equilibrium point.
solution
(a) Take PE at lowest point to be zero
ΔPE = mg Δ h = Fg Δ h
= (12 N)(-0.40m)
= -4.8 J
(b)
ΔKE = -ΔPE
= -(-4.8 J)
= 4.8 J
Sol’n cont’d
( c ) First must calculate
mass of bob
Fg = mg
m = Fg / g
= 12N / 9.8m/s2
= 1.2 kg
Then calculate velocity
KE = ½ mv2
v = sqrt (2KE /m)
= sqrt [ 2(4.8J)/(1.2kg) ]
= 2.8 m/s
Elastic Potential Energy and
Springs
Hooke’s Law
• The English scientist Robert Hooke was able to
show that the magnitude of a force (F) is directly
proportional to the elongation (stretch) of
compression of a spring (x) within certain limits.
Fs = kx
k – spring constant – unit is the newton per meter
(N/m)
Note: the greater the constant, the stiffer the spring
Elastic Potential Energy
Spring is attached to a wall. If
a force is applied and stretch
the string to the right, work
has been done. This work has
been converted into the
spring’s potential energy
– Elastic Potential Energy
PEs = ½ kx2
Hooke’s law and Potential Energy
Area under the graph
equals the work done.
W
Example
• A spring whose constant is 2.0 N/m is
stretched 0.40 m from its equilibrium position.
What is the increase in the elastic potential
energy of the spring?
Solution
PEs = ½ kx2
= ½ (2.0 N/m ) (0.40 m)2
= 0.16 J
Elastic and Inelastic Collisions
• In an elastic collision BOTH kinetic energy and
momentum are conserved
p1i + p2i = p1f + p2f
KE1i + KE2i = KE1f +KE2f
p1i
p2i
1
p1f
2
KE1i
KE2i
1
2
p2f
1
KE1f
2
KE2f
Inelastic collisions
• In inelastic collisions, the kinetic energy that is
“lost” is converted into internal energy (Q) of the
objects by frictional forces.
• These systems are called nonideal mechanical
systems.
• The energy is constant
ET = PE + KE + Q
A change in the internal energy of an object is
usually accompanied by a change in temperature
Simple Machines and Work
Simple Machines
• A simple machine is a device that allows work to be
done and offers and advantage to the user. Ex: pulleys,
levers, inclined planes, wheels and axles and
screwdrivers
Win = Wout
Fin ∙ din = Fout ∙ dout
Fout / Fin = din / dout
Mechanical Advantage (MA): Fout / Fin
Ideal MA (IMA) assumes no friction (use din / dout )
Actual MA (AMA) is always less than IMA (use Fout / Fin )
The efficiency of machine is AMA/IMA ratio and this
value is always less than 100%
example
• A 100 N object is moved 2 m up an inclined
plane whose end is lifted 0.5 m from the floor.
If a force of 50 N is needed to accomplish this
task, calculate the (a) IMA
(b) AMA
And
(c) efficiency of the inclined
plane
solution
Input force (Fin ) = 50 N (force needed to move object)
Output force (Fout ) = 100N (force that has been lifted)
Input distance (din ) = 2m (distance moved along plane)
Output distance (dout ) = 0.5m (distance object is raised)
a) IMA = din / dout = 2m / 0.5m = 4
b) AMA = Fout / Fin = 100N / 50 N = 2
c) efficiency = AMA/IMA = 2/4 = 0.5 (50%)
Internal Energy and Work
Internal Energy (Q)
• Internal Energy of a system is the total kinetic
and potential energies of the atoms and
molecules that make up the system.
• Recall: a change in the internal energy of an
object is usually accompanied by a change in
its temperature.
example
• Force is used to move an object along a
horizontal table at constant speed.
• Has work been done? Yes W = F ∙ d
• Is there a change in Kinetic Energy? Why or
why not? No – speed is constant
• Is there a change in Potential Energy? Why or
why not? No – table is horizontal - so no change in height
• How was work used? Used to overcome friction between
object and table so internal energy
of the object-table system has been
increased by work done.
The Laws of Thermodynamics
The study of the relationships among
heat, work, and energy in the
universe
The first law of Thermodynamics
• Energy can neither be created nor destroyed.
It can only change forms.
• The change in the internal energy of a system
(ΔU) is equal to the heat (Q) that the system
absorbs (or releases) minus the work (W) it
does (or has done on it)
ΔU = Q – W
2nd Law of Thermodynamics
• Result if the work of the French physicist
Nicolas Carnot with heat engines.
• Law states heat cannot flow from colder
object to a warmer one without work being
done on the system.
– Ex. Refrigerators must be run by motors in order
to withdraw heat from objects placed in them
• No heat engine can be 100% efficient. Some
of the heat absorbed by the engine must be
lost in the random motion of its molecules.
• Entropy is the measure of this disorder
3rd law of Thermodynamics
• As temperature approaches absolute zero, (0 K)
the entropy of a system approaches a constant
minimum
• The efficiency of a heat engine depends on its
operating temperatures; engine would reach
100 % efficiency only at 0 K.
– Engine cannot be completely efficient, therefore 0K
cannot be reached.