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SPH4U: Introduction to Work

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

Forms of Energy

Kinetic: Energy of motion.

A car on the highway has kinetic energy.

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:

+ 5,000,000,000 V

e-

- 5,000,000,000 V

e+ (a)

(b)

(c)

E = 1010 eV

M E = MC2

( poof ! )

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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”...

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

F

r Fr

“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)

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) Definition:

a.b = ab cos

= a[b cos ] = aba

= b[a cos ] = bab

Some properties:

ab = ba

q(ab) = (qb)a = b(qa) (q is a scalar) a(b + c) = (ab) + (ac) (c is a vector)

The dot product of perpendicular vectors is 0 !!

a

ab b

a

b

ba

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Work & Energy Understanding

A box is pulled up a rough (m > 0) incline by a rope-pulley-weight arrangement as shown below.

How many forces are doing work on the box?

(a) 2

(b) 3

(c) 4

Solution

Draw FBD of box: N

f

mg

T

Consider direction of motion of the box

v

Any force not perpendicular to the motion will do work:

N does no work (perp. to v)

T does positive work

f does negative work

mg does negative work

3 forces do work

Work: Example

(constant force)

A force F = 10 N pushes a box across a frictionless

floor for a distance x = 5 m.

x

F

Work done by F on box is:

WF = Fx = F x (since F is parallel to x)

WF = (10 N) x (5 m) = 50 Joules (J)

Units of Work:

Force x Distance = Work

N-m (Joule) Dyne-cm (erg)

= 10-7 J

BTU = 1054 J

calorie = 4.184 J

foot-lb = 1.356 J

eV = 1.6x10-19 J

cgs other mks

Newton x

[M][L] / [T]2

Meter = Joule

[L] [M][L]2 / [T]2

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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.

x

F

v1 v2

m

Work & Kinetic Energy...

Since the force F is constant, acceleration a will be

constant. We have shown that for constant a:

v22 - v1

2 = 2a(x2-x1) = 2ax.

multiply by 1/2m: 1/2mv22 - 1/2mv1

2 = max

But F = ma 1/2mv22 - 1/2mv1

2 = Fx

x

F

v1 v2

a m

Work & Kinetic Energy...

So we find that

1/2mv2

2 - 1/2mv12 = Fx = WF

Define Kinetic Energy K: K = 1/2mv2

K2 - K1 = WF

WF = K (Work/kinetic energy theorem)

x

F a m

v2 v1

Work/Kinetic Energy Theorem:

{Net Work done on object}

=

{change in kinetic energy of object}

netW K

f iK K

2 21 1

2 2f imv mv

University will prove this for a variable force later.

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Nice to Know

B

ABA

W dxF

B

ABA

maW dx 2 21 1

2 2AB B AW mv mv

21

2

B

A

v

AB

v

W mv

B

ABA

dvm

dtW dx

B

ABA

dvm vdt

dtW

B

AvAB

v

mvdvW

AB B AW KE KE

ABW KE

A

B

The work in going

from A to B.

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 (c) they will go the same distance

m1

m2

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).

m f

N

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.

m

D

This work “removes” the kinetic energy that the box had:

WNET = K2 - K1 = 0 - K1

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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 m2D2 m1D1

m1

D1

m2

D2

Since m1 > m2 we can see that D2 > D1

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

Wg = mgH

Work/Kinetic Energy Theorem:

Wg = mgH = 1/2mv2

r mg

H

j

v0 = 0

v 2v gH

What about multiple forces?

Suppose FNET = F1 + F2 and the

displacement is r.

The work done by each force is:

W1 = F1 r W2 = F2 r

WTOT = W1 + W2

= F1 r + F2 r

= (F1 + F2 ) r

WTOT = FTOT r It’s the total force that matters!!

FNET

r F1

F2

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

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Comments...

W = F r

No work done if = 90o.

No work done by T.

No work done by N.

T

v

v

N

Work & Energy Question

An inclined plane is accelerating with constant acceleration a.

A box resting on the plane is held in place by static friction.

How many forces are doing work on the block?

a

(a) 1 (b) 2 (c) 3

Solution

First, draw all the forces in the system:

a

mg N

FS

a

mg N

Solution

Recall that W = F Δr so only forces that have a component along the direction of the

displacement are doing work.

FS

The answer is (b) 2.

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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)

WorkPower

time

WP

t

J

s

Watt (W)

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:

av av

W FdP Fv

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?

WP

t

Fd

t

300 6

20

90

N m

s

W

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?

WP

t

t

Fd vF vmg

1000 9.8 8.0

78,000

78

N mkg

kg s

W

kW

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Flash