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SPH3U:
Introduction to Work and Energy
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Work & Energy
 Discussion
 Definition
Dot Product
Work of a constant force
 Work/kinetic energy theorem
Work of multiple constant forces
Comments
Work & Energy

One of the most important concepts in physics
 Alternative approach to mechanics

Many applications beyond mechanics
 Thermodynamics (movement of heat)
 Quantum mechanics...

Very useful tools
 You will learn new (sometimes much easier) ways to
solve problems
Energy is..
• The ability to do work
• Measured in Joules
• We use energy to do
whatever task we
need to do (move a
car, lift a pencil, etc)
Types of Energy
Energy can come in many forms including:
– Thermal (energy in the form of moving atoms)
– Electrical (energy possessed by charged particles)
– Radiant (energy found in EM waves)
– Nuclear (energy stored in holding the atom together)
– Gravitational (energy stored due to a raised elevation)
Types of Energy cont.
– Kinetic (energy due to motion of objects)
– Elastic (energy stored in compression or stretch)
– Sound (energy in vibrations)
– Chemical (energy stored in molecular bonds)
Energy 101
Energy cannot be created or destroyed, it can
only transform from one form to another.
• Eg. A student turns on the stove to heat a pot
of water
Electrical  Radiant  Thermal
Energy and Work
• We know energy is the ability to do work, but
what is work?
• In physics work is the energy transferred to an
object by an applied force over a displacement
Forms of Energy

Kinetic: Energy of motion.
 A car on the highway has kinetic energy.
1 2
K  mv
2



We have to remove this energy to stop it.
The brakes of a car get HOT!
This is an example of turning one form of energy into
another (thermal energy).
Mass = Energy

Particle Physics:
E = 1010 eV
(a)
e+
e+ 5,000,000,000 V
(b)
(c)
- 5,000,000,000 V
M
E = MC2
( poof ! )
Energy Conservation

Energy cannot be destroyed or created.
 Just changed from one form to another.

We say energy is conserved!
 True for any closed system.
 i.e. when we put on the brakes, the kinetic energy of the car is
turned into heat using friction in the brakes. The total energy of
the “car-brakes-road-atmosphere” system is the same.
 The energy of the car “alone” is not conserved...
 It is reduced by the braking.

Doing “work” on an isolated system will change its “energy”...
Work
• Mechanical work is done on an object when a force displaces an
object.
W=FΔd
• Note that this equation only applies when the force is constant and
the force and displacement are in the same direction.
• When the force and displacement are not entirely in the same
direction, the component of the force in the direction of the
displacement is used.
Definition of Work:
Ingredients: Force (F), displacement (r)
Work, W, of a constant force F
acting through a displacement r is:
W = F r = F r cos  = Fr r
“Dot Product”
The dot product allows us
to multiply two vectors, but
just the components that
are going in the same
direction (usually along
the second vector)
F

Fr
r
Definition of Work...

Only the component of F along the displacement
is doing work.
 Example: Train on a track.
F

r
F cos 
Aside: Dot Product (or Scalar
Product)
a
Definition:
.
a b = ab cos 
= a[b cos ] = aba
ba

b
a
= b[a cos ] = bab
Some properties:
ab = ba
q(ab) = (qb)a = b(qa)
a(b + c) = (ab) + (ac)

ab
(q is a scalar)
(c is a vector)
The dot product of perpendicular vectors is 0 !!
b
Work: Example
(constant force)

A force F = 10 N pushes a box across a frictionless
floor for a distance x = 5 m.
F
x
Work done by F on box is:
WF = Fx = F x
(since F is parallel to x)
WF = (10 N) x (5 m) = 50 Joules (J)
Units of Work:
Force x Distance = Work
Newton x Meter = Joule
[M][L] / [T]2 [L]
[M][L]2 / [T]2
mks
N-m (Joule)
cgs
Dyne-cm (erg)
= 10-7 J
other
BTU
calorie
foot-lb
eV
= 1054 J
= 4.184 J
= 1.356 J
= 1.6x10-19 J
Example 1
How much mechanical work will be done
pushing a shopping cart 3.5m with a force of
25N in the same direction as the displacement?
W  Fd
  25 N  3.5m 
 87.5 J
Example 2
A curler applies a force of 15.0N on a curling stone and
accelerates the stone from rest to a speed of 8.00m/s
in 3.5s. Assuming the ice surface is level and
frictionless, how much mechanical work is done on the
stone?
W  Fd
 15 N  d
 v0  v f 
d 
t
2


m
 m
0

8
 s

s

  3.5s 
2




 14 m
W  Fd
 15 N 14 m 
 210 J
Example 3
Calculate the work done by a custodian on a vacuum
cleaner if the custodian exerts an applied force of
50.0N on the vacuum hose and the hose makes a 30°
angle with the floor. The vacuum moves 3.0m to the
right on a flat level surface.
W  F d
 Fd cos  
  50 N  3m  cos  30 
 129.9 J
Useless work?
• No work: when there is no displacement, no
work is done! Work can also be positive or
negative relative to the motion, as shown in
the next example.
Comments:

Time interval not relevant
 Run up the stairs quickly or slowly...same Work
Since W = F r

No work is done if:
 F=0
or
 r = 0
or
  = 90o
Example 4
A shopper pushes a shopping cart on a horizontal
surface with a horizontal applied force of 41.0N for
11.0m. The cart experiences a force of friction of 35.0N.
Calculate the total work done on the cart.
W friction  F  d
Wshopper  F  d
  41 N 11m 
  35 N 11m 
 451 J
 385 J
WNet  Wshopper  W friction
 451 J  385 J
 66 J
Work & Kinetic Energy:

A force F = 10 N pushes a box across a frictionless
floor for a distance x = 5 m. The speed of the box is v1
before the push and v2 after the push.
v1
v2
F
x
m
Work & Kinetic Energy...

Since the force F is constant, acceleration a will be
constant. We have shown that for constant a:
 v22 - v12 = 2a(x2-x1) = 2ax.
1/ mv 2 - 1/ mv 2 = max
 multiply by 1/2m:
2
2
2
1
1/ mv 2 - 1/ mv 2 = Fx
 But F = ma
2
2
2
1
v1
v2
F
x
m
a
Work & Kinetic Energy...

So we find that
 1/2mv22 - 1/2mv12 = Fx = WF

Define Kinetic Energy K: K = 1/2mv2
 K2 - K1 = WF
 WF = K
(Work/kinetic energy theorem)
v2
v1
F
x
m
a
Work/Kinetic Energy Theorem:
{Net Work done on object}
=
{change in kinetic energy of object}
W net   K
 K f  Ki

1
1
mv f 2  mvi 2
2
2
Example

A 200g hockey puck initially at rest on ice is pushed by a
hockey stick by a constant force of 6.0N. What is the
hockey puck’s speed after it has moved 2m?
WW  KE
1 2 1 2
FNet d  mv f  mvi
2
2
 6.0 N 2m  
1
1
 m
2
0
.
2
kg
v

0
.
2
kg

 f 
0 
2
2
 2
12 J   0.1kg  v2f
2
2
m
v2f  120 2
s
m
 10.95
s
m
 11
s
Work & Energy Question

Two blocks have masses m1 and m2, where m1 > m2. They
are sliding on a frictionless floor and have the same kinetic
energy when they encounter a long rough stretch (i.e. m > 0)
which slows them down to a stop.
Which one will go farther before stopping?
(a) m1 (b) m2
m1
m2
(c) they will go the same distance
Solution


The work-energy theorem says that for any object WNET = K
In this example the only force that does work is friction (since
both N and mg are perpendicular to the block’s motion).
N
f
m
mg
Solution



The work-energy theorem says that for any object WNET = K
In this example the only force that does work is friction (since
both N and mg are perpendicular to the blocks motion).
The net work done to stop the box is - fD = -mmgD.


This work “removes” the kinetic energy that the box had:
WNET = K2 - K1 = 0 - K1
m
D
Solution


The net work done to stop a box is - fD = -mmgD.
 This work “removes” the kinetic energy that the box had:
 WNET = K2 - K1 = 0 - K1
This is the same for both boxes (same starting kinetic energy).
mm2gD2  mm1gD1
Since m1 > m2 we can see that
m2D2  m1D1
D2 > D1
m1
D1
m2
D2
A Simple Application:
Work done by gravity on a falling object


What is the speed of an object after falling a distance H, assuming it
starts at rest?
Wg = F r = mg r cos(0) = mgH
v0 = 0
Wg = mgH
r
mg
j
H
Work/Kinetic Energy Theorem:
Wg = mgH = 1/2mv2
v  2 gH
v
POWER
Simply put, power is the rate at which work gets done (or
energy gets transferred).
Suppose you and I each do 1000J of work, but I do the
work in 2 minutes while you do it in 1 minute. We both did
the same amount of work, but you did it more quickly (you
were more powerful)
Work
Power 
time
Watt (W)
P
W
t
J
s
• Power = change in energy
time
Measured in Watts (J/s or W) or in horsepower
Ex. 1 horsepower = 746 W or 0.75 kW
Vehicle horsepower can be calculated using its RPM & Torque (in the manual).
Watt determined that the average horse could do 33, 000 foot-pounds of work per
minute (Ex. Move 1000 lbs of coal 33 ft every minute).
The equation for power is:
P = ΔE = W
Δt
Δt
Power is a measure of:
the change in energy
over time.
aka.
Energy used or Work done
over time
Kahn Academy: Speed of Weight-Lifter:
https://www.youtube.com/watch?v=RpbxIG5HTf4
Power
Power is also needed for acceleration and for moving against the force of
gravity.
The average power can be written in terms of the force and the average
velocity:
W Fd
Pav 

 Fvav
t
t
Understanding
A mover pushes a large crate (m= 75 kg) from one side of a truck to the
other side ( a distance of 6 m), exerting a steady push of 300 N. If she
moves the crate in 20 s, what is the power output during this move?
P
W
Fd

t
t
300 N  6m 


20 s
 90W
Understanding
What must the power output of an elevator motor be such that it
can lift a total mass of 1000 kg, while maintaining a constant
speed of 8.0 m/s?
Fd
W

P
 Fv  mgv
t
t

N 
m
 1000kg   9.8  8.0 
kg  
s

 78,000W
 78kW