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Mechanical Work: Learning Goals
The student will be
able to explain the
concepts of and
interrelationships
between energy,
work, and power, and
identify and describe
their related units
(D3.2).
Mechanical Work
The mechanical work is the amount of mechanical
energy transferred to an object by a force.
Mechanical energy of an object = kinetic energy
+ potential energies
Mechanical Energy
The mechanical energy of an object is that part of
its total energy which is subject to change by
mechanical work.
(definition by the Department of Redundancy Department)
This means energy is a concept so fundamental in
physics that it is not easily defined in terms of
anything more fundamental.
Mechanical Work and Energy
Therefore work is a transfer of mechanical energy
and vice versa:
𝑊 = ∆𝐸
Energy allows work to be done. When work is done
it changes the energy of a system.
Examples of Work: (no, not your part-time job!)
❖ throwing a baseball
❖ hitting a hockey puck
❖ kicking a football
❖ shovelling snow
❖ lifting weights
❖ running a marathon
❖ ???
Mechanical Work
The amount of work depends on:
- the magnitude of the force
- displacement of the object
Equation time….!!!
Get Back to Work Already!!!
The simplest case of work done an object moving in
1D with a constant force parallel to that direction is:
𝑊 = 𝐹∆𝑑where W = work done (J)
F is the force (N)
Dd is the displacement of the object (m)
Note: W is a scalar (no direction). Both F and Dd
are vectors; the dot product yields a scalar, W.
Unit Analysis
The SI unit of work and energy is Joule:
1 𝐽 = 1 𝑁 × 1 𝑚
1 𝐽 = 1 𝑘𝑔 ×𝑚
𝑠2× 1 𝑚
1 𝐽 = 1 𝑘𝑔 ×𝑚2
𝑠2 James Joule
Example 1
Captain America shoves Ultron
(m = 243 kg) horizontally with 40.0 N
of force. If Ultron moves 15.0 m, find
the work done on Ultron.
W
F N
d m
=
=
=
?
.
.
40 0
150D
W F d
W N m
W J
=
=
=
D
( . )( . )
.
40 0 150
6 00 102
Work: A More Complete Equation
The only portion of the force that contributes towards
the work done is the component of the force in the
direction of the motion:
𝑊 = 𝐹𝑐𝑜𝑠𝜃∆𝑑
The x-component of the
force (Fcos𝜃) is parallel to Dd.
Work: A More Complete Equation
When F is II to ∆𝑑, the equation simplifies to a
special case:
𝑊 = 𝐹𝑐𝑜𝑠𝜃∆𝑑𝑊 = 𝐹𝑐𝑜𝑠 0°∆𝑑𝑊 = 𝐹 1 ∆𝑑∴ 𝑊 = 𝐹∆𝑑
Example 2
Ultron pushes Captain America back
with 40.0 N at 60.0o to the horizontal.
If Captain America stumbles 15.0 m, find
how much work did Ultron do?
W
F N
d m
=
=
=
=
?
.
.
.
40 0
150
60 0
D
JW
mNW
dFW
21000.3
)0.15)(0.60)(cos0.40(
)cos(
=
=
D=
Work: Individual Forces
Work is calculated for each individual force not Fnet.
If there is an applied force, it will do positive work on
the object and total mechanical energy (M.E.) ↑.
If there is a frictional force opposing motion, it will do
negative work on the object and total M.E. ↓.
Positive and Negative Work
• Positive work: F and Dd are both in same direction, this increases an object’s energy
• Negative work: F and Dd are in opposite directions, this reduces an object’s energy
• Calculation of work corresponds to direction:
Example 3
Iron Man pushes Ultron 15.0 m across
the floor. The force of friction is 10.0 N.
Find the work done on Ultron by friction
causing him to stop.
Energy is removed as sound and heat - ouch!
Note: F and Dd act in opposite directions.
JW
mNW
dFW
21050.1
)0.15)(180)(cos0.10(
)cos(
−=
=
D=
When No Work is Done on an
Object: Case 1When a force exists but an object does not move,
𝑊 = 0.
Ex. A student pushes desperately with all their might
against the wall of the school hoping to topple it .
Because the school does not move ∆𝑑 = 0:
𝑊 = 𝐹∆𝑑𝑊 = 𝐹 0∴ 𝑊 = 0
Note: Energy is expended by the student but it does
not do any work on the school building.
When No Work is Done on an
Object: Case 2
When an object is in uniform motion but does not
require a force to keep it in motion, 𝑊 = 0.
Ex. A hockey puck slides across essentially a
frictionless ice rink; a satellite is launched from
Earth and enters space. Initially there was a force
applied but now that 𝐹 = 0:
𝑊 = 𝐹∆𝑑𝑊 = 0 ∆𝑑∴ 𝑊 = 0
When No Work is Done on an
Object: Case 3
When a force acting on an object is perpendicular to
the direction of motion, 𝑊 = 0.
Ex: Walking to class while holding your textbook in
front of your body:
𝑊 = 𝐹𝑐𝑜𝑠𝜃∆𝑑𝑊 = 𝐹𝑐𝑜𝑠90°∆𝑑
∴ 𝑊 = 0
Note: Initially work must be done to lift the book, but
then work on the book is zero as you walk forward.
Example 4The Hulk applies a horizontal
force of 40.0 N at 90° to carry the
massive 243 kg Ultron 20.0 m [E].
Find the work done on Ultron.
𝑊 = 0Hulk, who is ever mighty, is
applying a force at a right angle to
the motion which does not change
Ultron’s mechanical energy.