Corinne Goldstein,Ali Lubin,

Bridget Miller,and

Stephen Desiletspresent…

In association with DreamWORKS studios…

A Chapter 6 Production…

…And Energy

What’s this all about?

Well, let’s look at the terms…

Work – Measured in Joules

Energy – Also measured in Joules

Spring Constant – Measured in Newton/meters

Power – Measure in Watts

Well Work and Energy are both measured in

Joules.

Are they the same thing?

NO

Work (W)

• Work is the Transfer of Energy, or more simply put, Force exerted over a Distance

• W = FdcosӨ– F = Magnitude of Force (Newtons)– d = Magnitude of Displacement– Ө = Angle between Force and Displacement

So 1 Joule = 1 Newton x 1 Meter

And if someone walks down a horizontal road…

Work is done by that person…

But the work done by gravitational force is zero since the

displacement is perpendicular to the direction of the force

HOWEVER…

Energy of a body is its Capacity to do

Work

EnergyThere’s 3 Types!

Kinetic Energy (KE or K)

• The Energy of Motion

• KE = ½ mv²– m = mass of the object

– v = velocity of the object

Gravitational Potential Energy

• The stored energy an object has due to its position above earth’s surface

• PE = mgh– m = mass of the object– g = acceleration of gravity– h = height of the object relative to earth (or

some other zero level)

Elastic Potential Energy (PEs)

• Energy a Spring has by virtue of being stretched or compressed

• PEs = ½ kx²– k = Spring Constant – Measures the strength

of the spring in N/m– x = Distance the spring is stretched or

compressed

Say, are Work and Energy related?

YES

Work exerted on an object is equal to the difference between

an object’s initial and final Energies.

So…

WKE = KE – KE0 = mv² - ½ mv0²

Wgravity = PE – PE0 = mgh – mgh0

ormg(h-h0)

WPEs = PEs – PEs0 = ½ kx² - ½ kx0²

One of those equations even has its own

theorem!

Can you guess which one?

Work-Energy TheoremWKE = KE – KE0 = mv² - ½ mv0²

Or in English terms…

The work done by an object by an outside force is equal to the difference between the

object’s final and initial kinetic energies

So have I mastered the Force?

Not yet.Much to learn there still is,

Young Padawan.

Like the difference between Conservative Forces and Nonconservative Forces

Conservative Forces

• Force that exerts the same amount of force on an object moving between two points, no matter what path the object takes

• Types:– Gravitational Potential Energy– Elastic Spring Force– Electric Force

What’s all that Mumbo Jumbo mean, though?

Let’s think about it.

Take the Wgravity for example…

The equation is Wgravity = mg(h-h0)

So if an object falls 50 meters, the work exerted on the object by gravity won’t change whether the object fall straight

down or falls on an angle. All that matters is the initial and final heights.

Well then what’s a Non-Conservative Force?

Non-Conservative Force

• Force that exerts a different amount of work on an object moving between two points depending on what path the object takes

• Types:– Static and Kinetic Frictional Forces– Air Resistance– Tension– Normal Forces– Propulsion Force of a Rocket

Okay, well what if I want to calculate the work due to non-

conservative forces acting on an object?

Good Question!

Effectively, you can calculate the work exerted on an object by

non-conservative forces by…

finding the difference between the initial and final energies due to

conservative forces acting on an object.

The difference is the work due to non-conservative forces that acts on

the object

So…

Wnc = ΔKE + ΔPE + Δ PEs

orWnc = ( ½mv² - ½ mv0²) + (mgh – mgh0) + (½ kx² - ½kx0²)

(Δ means change)

Let’s Conserve Energy!

Mechanical Energy (ME or E) of and object is the sum of the

kinetic and potential energies of that object

E = KE + PE

When the net work exerted on a moving object by non-

conservative forces is zero, the total mechanical energy of that

object remains constant…

…In this case, KE and PE can be transformed into one another

as the object moves.

Cool Beans!

If there are no non-conservative forces acting on the object, then the initial and final total energies

should be equal

So…

Wnc = (½mv² + mgh + ½kx²) - (½mv0² + mgh0 + ½kx0²)or

½mv² + mgh + ½kx² = ½mv0² + mgh0 + ½kx0²

I GOT THE POWER!

Power

• The Rate at which Work is done• P = W/t or• P = Fv

or • P = (FdcosӨ)/t

• Measured in Watts (W)

So…

1 Watt = 1 Joule per second

We’re Almost Done!

Hey, remember howWork = Force x Distance?

Well, if you plot Force on one axis of a graph and Distance on

the other, then you can calculate the Work exerted by finding the area under the line.

Just likeArea = Length x Width,

Work = Force x Distance

Time to Practice!

Find the work exerted on a suitcase if the suitcase is being pulled with a force of 16.0 N at a 56.0º angle to the floor and the displacement of the suitcase is

63.0 cm.

W = FdcosӨW = (16.0N)(0.63m)(cos56.0º)

Degree Mode is your friend.

W = 5.64 J

An archer pulls a bowstring back a distance of 0.470m and then releases the arrow. The

bow and string act like a spring whose spring constant is 425

N/m. What is the elastic potential energy of the system?

PE = ½kx²PE = ½(425N/m)(0.470m)²

PE = 46.9 J

A motorcyclist is trying to leap across a canyon by driving

horizontally off the cliff at a speed of 38.0 m/s. Ignoring air

resistance, find the speed at which the motorcycle strikes the ground on the other side. The motorcycle starts at a height of 70.0m and will

end at a height of 35.0m.

½ mv² + mgh = ½mv0² + mgh0

(mass cancels out)

½v² + (9.8m/s²)(35.0m) = ½(38.0m/s)² + (9.8m/s²)(70.0m)

V = 46.2 m/s

A 0.20kg rocket is launched from rest. It takes a roundabout route until it reaches a height of 29m above its

starting point. In the process, 425J of work is done on the rocket by non-conservative forces (the burning

propellant). Ignoring air resistance and the mass lost due to the burning of the fuel, find the speed of the rocket when

it is 29 m above its starting point.

Wnc = ½mv² + mgh – (½mv0² + mgh0) 425J = [½(0.20kg)v² + (0.20kg)(9.8m/s²)(27m)] – [½(0.20kg)(0m/s)² + (0.20kg)(9.8m/s²)(0m)]

V = 61 m/s

A car starts from rest and accelerates in a positive

direction. The car has a mass of 1.10 x 10³kg and accelerates

at +4.60m/s² for 5.00s. Determine the average power

generated by the force that moves the car.

V = v0 + atv = 0m/s + (4.6m/s)(5.00s)

v = 23 m/s

P = FvP = mav

P = (1.10 x 10³kg)(4.6m/s²)(23m/s)P = 1.16 x 10^5 W

Th-Th-That’s All Folks!

It’s Physics-tastic!