SPH3U: Introduction to Work and Energy 

   

Work & Energy

Work & Energy  Discussion  Definition Dot Product Work of a constant force  Work/kinetic energy theorem Work of multiple constant forces Comments

Energy is..



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

Types of Energy

• The ability to do work Energy can come in many forms including:

• Measured in Joules

– Thermal (energy in the form of moving atoms)

• We use energy to do

– Electrical (energy possessed by charged particles)

whatever task we

– Radiant (energy found in EM waves)

need to do (move a

– Nuclear (energy stored in holding the atom together)

car, lift a pencil, etc)

– Gravitational (energy stored due to a raised elevation)

Types of Energy cont.

Energy 101

– Kinetic (energy due to motion of objects)

Energy cannot be created or destroyed, it can

– Elastic (energy stored in compression or stretch)

only transform from one form to another.

– Sound (energy in vibrations) – Chemical (energy stored in molecular bonds)

• Eg. A student turns on the stove to heat a pot of water Electrical  Radiant  Thermal

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Energy and Work

Forms of Energy 

Kinetic: Energy of motion.  A car on the highway has kinetic energy.

• We know energy is the ability to do work, but

1 K  mv2 2

what is work? • In physics work is the energy transferred to an

 

object by an applied force over a displacement



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 

Energy Conservation

E = 1010 eV (a)



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”...

Particle Physics: e+

e+ 5,000,000,000 V

(b)

- 5,000,000,000 V M

E = MC2

( poof ! )

(c)

Work

Definition of Work:

• Mechanical work is done on an object when a force displaces an

Ingredients: Force (F), displacement (r)

object.

Work, W, of a constant force F acting through a displacement r is: W = F r = F r cos  = Fr r

W=FΔd • Note that this equation only applies when the force is constant and

“Dot Product”

the force and displacement are in the same direction. • When the force and displacement are not entirely in the same

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)

direction, the component of the force in the direction of the displacement is used.

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F 

Fr

r

Aside: Dot Product (or Scalar Product)

Definition of Work...

a

Definition: 

Only the component of F along the displacement is doing work.  Example: Train on a track.

ba

.

a b = ab cos 



= a[b cos ] = aba

b a

F 

= b[a cos ] = bab r



Some properties: ab = ba q(ab) = (qb)a = b(qa) a(b + c) = (ab) + (ac)

F cos 

b

ab (q is a scalar) (c is a vector)

The dot product of perpendicular vectors is 0 !!

Work: Example (constant force) 

Units of Work:

A force F = 10 N pushes a box across a frictionless floor for a distance x = 5 m.

Force x Distance = Work Newton x Meter = Joule [M][L] / [T]2 [L] [M][L]2 / [T]2

F mks x

N-m (Joule)

Work done by F on box is:

cgs Dyne-cm (erg) = 10-7 J

WF = Fx = F x (since F is parallel to x) WF = (10 N) x (5 m) = 50 Joules (J)

other BTU calorie foot-lb eV

= 1054 J = 4.184 J = 1.356 J = 1.6x10-19 J

Example 2

Example 1

A curler applies a force of 15.0N on a curling stone and

How much mechanical work will be done

accelerates the stone from rest to a speed of 8.00m/s

pushing a shopping cart 3.5m with a force of

in 3.5s. Assuming the ice surface is level and

25N in the same direction as the displacement?

frictionless, how much mechanical work is done on the stone?

W  Fd

W  Fd

  25 N  3.5m 

 15 N  d

 87.5 J

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 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

Useless work?

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°

• No work: when there is no displacement, no

angle with the floor. The vacuum moves 3.0m to the

work is done! Work can also be positive or

right on a flat level surface.

negative relative to the motion, as shown in

W  F d

the next example.

 Fd cos     50 N  3m  cos  30   129.9 J

Example 4

Comments:

A shopper pushes a shopping cart on a horizontal surface with a horizontal applied force of 41.0N for



Time interval not relevant  Run up the stairs quickly or slowly...same Work

11.0m. The cart experiences a force of friction of 35.0N. Calculate the total work done on the cart. W friction  F  d

Since W = F r 

Wshopper  F  d

No work is done if:  F=0 or  r = 0 or   = 90o

  41 N 11m 

  35 N 11m 

 451 J

 385 J

WNet  Wshopper  W friction  451 J  385 J  66 J

Work & Kinetic Energy: 

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

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

m

x

x

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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)

Work/Kinetic Energy Theorem: {Net Work done on object} = {change in kinetic energy of object}

W net   K  K f  Ki

v2

v1 F

m



a

1 1 mv f 2  mvi 2 2 2

x

Example 

Work & Energy Question

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

m2 s2 m  10.95 s m  11 s

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?

v2f  120

1 1 FNet d  mv2f  mvi2 2 2

 6.0 N 2m    0.2kg  v2f   0.2kg   0 1 2



m   2

1 2

12 J   0.1kg  v2f

2

(a) m1 (b) m2

m1

m2

Solution  

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).

  

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.  

N f

(c) they will go the same distance

This work “removes” the kinetic energy that the box had: WNET = K2 - K1 = 0 - K1

m

m

mg

D

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A Simple Application: Work done by gravity on a falling object

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

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 Since m1 > m2 we can see that

Work/Kinetic Energy Theorem:

D2 > D1

Wg = mgH = 1/2mv2 m1

v  2 gH

m2

D1

v

D2

POWER • Power = change in energy time

Simply put, power is the rate at which work gets done (or energy gets transferred).

Measured in Watts (J/s or W) or in horsepower

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)

Power  Watt (W)

P

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).

Work time

W t

The equation for power is:

J s

P = ΔE = W Δt Δt

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:

Power is a measure of:

Pav 

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

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W Fd   Fvav t t

Understanding

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?

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?

P

W Fd P  t t 300 N  6m    20 s  90W

Fd W   Fv  mgv t t  N  m  1000kg   9.8  8.0  kg   s   78,000W

 78kW

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