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.