november 04, 2016 - mr herman's webpagework energy presentation part 1 2016.notebook 3 november 04,...

Work Energy Presentation Part 1 2016.notebook

1

November 04, 2016

www.njctl.org

Algebra Based PhysicsWork and Energy

Table of Contents

Work and EnergyClick on the topic to go to that section

• Energy and the WorkEnergy Theorem

• Forces and Potential Energy

• Conservation of Energy

• Power

https://www.njctl.org/video/?v=zFxiRGMN5RE

Work and Work Energy Theorem

Return toTable ofContents

Work and the WorkEnergy Theorem

https://www.njctl.org/video/?v=gpIGt6ANfFU

Dec 76:03 PM

The most powerful concepts in science are called "conservation principles". These principles allow us to solve problems without worrying too much about the details of a process.

We just have to take a snapshot of a system initially and finally; by comparing those two snapshots we can learn a lot.

Conservation Principles

Dec 76:03 PM

If you know that there are 50 pieces of candy at the beginning. And you know that none of the pieces have been taken out or added...you know that there must be 50 pieces at the end.


A good example is a bag of candy.

Dec 76:03 PM


You can change the way you arrange them by moving them around...but you still will have 50 pieces.In that case we would

say that the number of pieces of candy is conserved.

That is, we should always get the same amount, regardless of how they are arranged.

https://www.njctl.org/video/?v=zFxiRGMN5REhttps://www.njctl.org/video/?v=gpIGt6ANfFU


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November 04, 2016

Dec 76:03 PM

Energy is a conserved property of nature. It is not created or destroyed. Therefore in a closed system we will always have the same amount of energy.

The only way the energy of a system can change is if it is open to the outside...this means that energy has been added or taken away.

Conservation of Energy

Dec 76:03 PM

We may not be able to define energy, but because it is a conserved property of nature, it's a very useful idea.

What is Energy?

It turns out that energy is so fundamental, like space and time, that there is no good answer to this question. However, just like space and time, that doesn't stop us from doing very useful calculations with energy.

Dec 76:03 PM

If we call the amount of energy that we start with "Eo" and the amount we end up with as "Ef" then we would say that if no energy is added to or taken away from a system that

Eo = EfIt turns out there are only two ways to change the energy of a system. One is with heat (which we won't deal with here) the

other is with Work, "W".

If we define positive work as that work which increases the energy of a system our equation becomes:

Eo + W = Ef


Dec 76:03 PM

Work can only be done to a system by an external force; a force from something that is not a part of the system.

Work

So if our system is a plane on an aircraft carrier and we come along and push the plane, we can increase the energy of the plane…

We are essentially doing work on the plane.

Dec 76:03 PM

The amount of work done, and therefore the amount of energy increase that the system will experience is given by the equation:

Work

There are some important points to understand about this equation.

W = Fdparallel

Meaning, work is the product of the force applied which moves the object a parallel displacement

Dec 76:03 PM

WorkIf the object that is experiencing the force does not move

(if dparallel = 0) then no work is done.

The energy of the system is unchanged; a state ofequilibrium.


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November 04, 2016

Jul 110:16 AM

Acceleration occurs due to the unbalanced force.Work is the ability to cause change.

Positive Work

DisplacementMF

If the object moves in the same direction as the direction of the force(for instance if force and displacement are in the same direction)

then the work is positive: W > 0.

The energy of the system is increased.

Jul 110:16 AM

If the object moves in the direction opposite the direction of the force (for instance if force and displacement are in opposite directions) then the work is negative: W


4

November 04, 2016

Two Dimensional Forces and Work

FAPP

Fparallelv Fperpendicular

Δx

θ

After breaking FAPP into components that are parallel and perpendicular to the direction of motion, we can see that no work is done by the perpendicular component; work is only done by the parallel component.


Using trigonometry, we find that Fparallel = FAPPcosθ


FAPP

Fparallelv Fperpendicular

Δx

θ

In words, the work done on an object by a force is the product of the magnitude of the force and the magnitude of the displacement times the cosine of the angle between them.


W = Fparalleld becomes:

W = (FAPPcosθ)Δx = FAPPΔxcosθ


This is really no more difficult a case. We just have to find the component of force that is parallel to the object's displacement.


Instead of pulling the objectat an angle to the horizontal,what if it is pushed?


The interpretation is the same, just determine the angle between the force and displacement and use:

W = FAPPΔxcosθ

Fparallel

FperpendicularθFAPP

Δx


Even though Fperpendicular is in the negative direction (it was positive when the object was pulled), it does not affect the work as only the parallel component contributes to the work.

Dec 76:03 PM

Eo + W = Ef

Since the work changed the energy of a system: the units of energy must be the same as the units of work

The units of both work and energy are the Joule.

Units of Work and Energy

James Joule

Dec 76:31 PM

1 A +24 N force is applied to an object that moves 10 m in the same direction during the time that the force is applied. How much work is done to the object?

Answ

er

https://www.njctl.org/video/?v=63FEFiw9qg

https://www.njctl.org/video/?v=63FEFi-w9qg


5

November 04, 2016

Dec 76:31 PM

2 A barbell of mass "m" is lifted vertically upwards, at a constant velocity, to a distance "h" by an outside force. How much work does that outside force do on the barbell?

A mgB mghC mghD 0E mg

Hint: Do a free body diagram to determine a formula for the outside force (Fapp); then use the formula for work: W = Fdparallel.

Answ

er

https://www.njctl.org/video/?v=F9K5YlPoZk

Gravitational Potential Energy


Return to Tableof Contents


We'll start with a quick review of Gravitational Potential Energy as presented in the Algebra Based Physics course. Then, the impact of motion in two dimensions will be analyzed.


Fapp

mg

A book of mass, m, is lifted vertically upwards a displacement, h, by an external force (a person) at a constant velocity. How much work does the external force do on the book?



Fapp

mg

a = 0Apply Newtons' Second Law to the FBD:

ΣFy = Fapp mg = may = 0 (since v = const)

Fapp = mg

Use the definition of work: W = FΔxparallel

W = (mg)Δxparallel = mgΔx (since Fapp is parallel to Δx)

W = mgh (since Δx = h)



Apply the WorkEnergy equation, Eo + W = Ef. If the book started and finished at rest (KE0 = KEf = 0), then Eo = 0. The equation becomes:

W = Ef

But we just showed that the external force did W = mgh to lift the book... so

mgh = Ef or Ef = mgh

We conclude that the energy of a mass is increased by an amount mgh when it is raised by a displacement, h.



The name for this form of energy is Gravitational Potential Energy (GPE).

GPE = mgh

One important thing to note is that changes in gravitational potential energy are important, but their absolute value is not. This is because h represents Δx, or the change in height.

You can define any point to be zero meters in height when it represents the original position of an object. This would give you zero for GPE at that point. But no matter what height you choose to call zero, changes in heights will still result in changes of GPE.

https://www.njctl.org/video/?v=F-9K5YlPoZk


6

November 04, 2016


Gravitational Potential EnergyThe book is moving up at a constant velocity and travels a displacement h. Note that the force exerted by gravity is in the opposite direction of its motion (antiparallel).

So, Wgrav = FgravΔxparallel

Wgrav = (mg)(Δx) = mgh

GPE = Wgrav = mgh

Which, of course, gives us the same result for GPE.

Fapp

mg

a = 0

Gravitational Potential Energy problem

3 What is the change of GPE for a 5.0 kg object which is raised from an initial height of 1.0 m above the floor to a final height of 8.0 m above the floor?

Answ

erGravitational Potential Energy problem

4 A librarian takes a book off of a high shelf and refiles it on a lower shelf. Which of the following explain the work done on the book by the librarian and the earth's gravitational field as its GPE is lowered? Select two answers.

A The librarian does positive work on the book.

B The librarian does negative work on the book.

C The gravitational field does positive work on the book.

D The gravitational field does negative work on the book. Ans

wer

B, C

GPE on an Incline

GPE on an Incline

h d

θ

Let's put together the concepts of two dimensional motion and forces with GPE.

We'll use a box being pushed up an incline.

It all depends on what we can measure. Assume it's easier to measure the displacement (d) the box travels.

How do we find its change in GPE?

GPE on an Incline

GPE on an Incline

h d

θ

The box starts with no velocity, and after it is pushed up a displacement d, the block slides up, and it momentarily stops before sliding back down.

Does ΔGPE = mgd?

GPE on an Incline

GPE on an Incline

h d

θ

No! The formula for ΔGPE was calculated from the work formula, and it assumed the gravitational force (or the force that opposed it to lift the object) was in the same direction of the object's motion.

The gravitational force points down. Since work only includes the distance and force components that are in parallel, ΔGPE involves h and not d in the picture.

How is h calculated from d?


7

November 04, 2016

GPE on an Incline

GPE on an Incline

h d

θ

ΔGPE = mgh

When motion is along an incline, the change in height can be related to the distance traveled using trigonometry.

sinθ = h/d

h = dsinθ

Mar 37:09 PM

5 What is the change in GPE for a 10.0 kg object which is raised from an initial height of 1.0 m above the floor to a final height of 10.0 m above the floor?

Answ

er

https://www.njctl.org/video/?v=HRfzm3aIJs

Mar 37:11 PM

6 What is the change in height of a 1/2 kg object which lost 20 J of GPE?

Answ

er

https://www.njctl.org/video/?v=m2Grf6ZpXcU

GPE on an Incline problem

7 A 5.0 kg block is at the top of a 6.0 m long frictionless ramp, which is at an angle of 370. What is the height of the ramp?

37o Answ

er

GPE on an Incline problem

8 The 5.0 kg block slides to the bottom of the 6.0 m long frictionless ramp, which is at an angle of 37o. What is the change in its GPE?

37o Ans

wer

Jul 710:06 AM

Answ

er

9 As an object falls downward, its GPE always _____.

A increases

B decreases

C stays the same

https://www.njctl.org/video/?v=5BOfc00McQs

https://www.njctl.org/video/?v=HRfzm-3aIJshttps://www.njctl.org/video/?v=m2Grf6ZpXcUhttps://www.njctl.org/video/?v=5BOfc00McQs


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November 04, 2016

Dec 76:03 PM

Kinetic Energy

Imagine an object of mass "m" at rest at a height "h". If dropped, how fast will it be traveling just before striking the ground?

Use your kinematics equations to get a formula for v2.

Since vo = 0, ∆x = h, and a = g

We can solve this for "gh"

We're going to use this result later.

v2 = vo2 + 2a∆x

v2 = 2gh

gh = v2 / 2

https://www.njctl.org/video/?v=AjT9BMKfze0

Dec 76:03 PM

Kinetic Energy

In this example, we dropped an object. While it was falling, its energy was constant...but changing forms.

It only had gravitational potential energy, GPE, at beginning, because it had height but no velocity.

Just before striking the ground (or in the example on the right, before hitting the hand) it only had kinetic energy, KE, as it had velocity but no height.

In between, it had some of both.

Dec 76:03 PM

Kinetic EnergyNow let's look at this from an energy perspective.

No external force acted on the system so its energy is constant. Its original energy was in the form of GPE, which is "mgh".

W = 0 and E0 = mgh

Solving for gh yields

Now let's use our result from kinematics(gh = v2/2)

This is the energy an object has by virtue of its motion: its kinetic energy

Eo + W = Ef

mgh=Ef

gh=Ef/m

v2/2=Ef/m

Ef=(1/2)mv2

Divide both sides by m

Dec 76:03 PM

Kinetic Energy

The energy an object has by virtue of its motion is called its kinetic energy. The symbol we will be using for kinetic energy is KE.

Like all forms of energy, it is measured in Joules (J).

The amount of KE an object has is given by:

KE = 1/2 mv2

Jul 710:05 AM

10 As an object falls, its KE always _____. A decreasesB increases

C stays the same.

Answ

er

https://www.njctl.org/video/?v=mcuU9nMfPdU

Jul 710:03 AM

11 A ball falls from the top of a building to the ground below. How does the kinetic energy (KE) compare to the potential energy (PE) at the top of the building?A KE = PE

B KE > PE

C KE


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November 04, 2016

Dec 76:49 PM

12 What is the kinetic energy of a 12 kg object with a velocity of 10 m/s?

Answ

er

https://www.njctl.org/video/?v=bdSb4Fqh9vg

Dec 89:08 AM

13 A 3 kg object has 45 J of kinetic energy. What is its velocity?

Answ

er

https://www.njctl.org/video/?v=XMqEio3puV0

Jul 710:20 AM

14 If the speed of a car is doubled, the KE of the car is:A quadrupled

B quartered

C halved

D doubled

Answ

er

https://www.njctl.org/video/?v=QuLGUEmsN7c

Jul 710:20 AM

15 If the speed of a car is halved, the KE of the car is:A quadrupled

B quartered

C halved

D doubled

Answ

er

https://www.njctl.org/video/?v=ISOCvADZWtc

Jul 710:21 AM

16 Which graph best represents the relationship between the KE and the velocity of an object accelerating in a straight line?

KE

v

KE

v

KE

v

KE

v

A

B

C

D Ans

wer

https://www.njctl.org/video/?v=c9drhr5pCY

Jul 710:35 AM

17 The data table below lists mass and speed for 4 objects. Which 2 have the same KE?

A A and D

B B and D

C A and C

D B and C

Answ

er

https://www.njctl.org/video/?v=Dw91asRRhSk

https://www.njctl.org/video/?v=bdSb4Fqh9vghttps://www.njctl.org/video/?v=XMqEio3puV0https://www.njctl.org/video/?v=QuLGUEmsN7chttps://www.njctl.org/video/?v=ISOCvADZWtchttps://www.njctl.org/video/?v=c9d-rhr5pCYhttps://www.njctl.org/video/?v=Dw91asRRhSk


10

November 04, 2016

Dec 1512:18 PM

Elastic Potential Energy

Energy can be stored in a spring, this energy is called Elastic Potential Energy.

Robert Hooke first observed the relationship between the force necessary to compress a spring and how much the spring was compressed.

https://www.njctl.org/video/?v=V7FjkmolSZk

Dec 1512:18 PM

Hooke's Law

Fspring = kx

k represents the spring constant and is measured in N/m.

x represents how much the spring is compressed and is measured as you would expect, in meters.

The sign tells us that this is a restorative force. (if you let the spring go once it is compressed, it

will go back to its original position)

Dec 1512:18 PM

Hooke's Law

Fspring = kx

F(N)

x (m)Force (effort required to stretch)

Displacement (elongation)

If we graph the relationship between force and elongation the mathematical relationship

can be experimentally confirmed.

Dec 1512:18 PM

Hooke's Law

Fspring = kx

Varying the displacement/elongation (x)

F(N)

x (m)small elongations require small forces

F(N)

x (m)large elongations require large forces

Dec 1512:18 PM

Hooke's LawFspring = kx

Varying the spring constant k (the stiffness of the spring)The spring constant is related to the slope the line.

F(N)

x (m)

Spring C

onstant = slope of lin

e (Newto

ns/meter)

Dec 1512:18 PM

Hooke's LawFspring = kx

Varying the spring constant k (the stiffness of the spring)The spring constant is related to the slope the line.

F(N)

small spri

ng constan

tlarge spring constant

x (m)



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November 04, 2016

Mar 1712:41 PM

18 Which spring requires a greater force to stretch?

A blueB greenC the same force is required

F(N)

x (m)

small spri

ng constan

tlarge spring constant

Answ

er


Mar 1712:36 PM

19 An ideal spring has a spring constant of 25N/m. Determine the force required to elongate/displace the spring by 2 meters.

Answ

er

https://www.njctl.org/video/?v=7tvI8d7Pd4A

Mar 1712:39 PM

20 A force of 100 newtons is applied to a spring with a constant of 25 N/m. Determine the resulting displacement/elongation.

Answ

er

https://www.njctl.org/video/?v=7tvI8d7Pd4A

Dec 1512:25 PM

Elastic Potential EnergyThe work needed to compress a spring is equal to the area

under its force vs. distance curve.

W = 1/2 (x)(F)

W = 1/2 (x)(kx)

W = 1/2kx2

Work = EPE

Area of a triangle = 1/2 b h

F = kx(N)

x (m)Area under curve is work done.

Area under curve is the elastic potential energy.

Area is in the shape of a triangle.

https://www.njctl.org/video/?v=nwaX5D0W1GU

Dec 1512:34 PM

Elastic Potential Energy

The energy imparted to the spring by this work must be stored in the Elastic Potential Energy (EPE) of the spring:

Like all forms of energy, it is measured in Joules (J).

EPE = 1/2 k x2

Dec 1512:25 PM

Elastic Potential EnergyWork done when varying the displacement/enlongation (x)

Remember the elastic potential energy stored is equal to the area under the curve.

F = kx(N)

x (m)

F = kx(N)

x (m)small elongation large elongation

small area

small EPE

large area

large EPE

EPE = 1/2 k x2Remember large elongations do large amounts of work.

https://www.njctl.org/video/?v=V7FjkmolSZkhttps://www.njctl.org/video/?v=7tvI8d7Pd4Ahttps://www.njctl.org/video/?v=7tvI8d7Pd4Ahttps://www.njctl.org/video/?v=nwaX5D0W1GU


12

November 04, 2016

Dec 1512:25 PM

Elastic Potential EnergyWork done when varying the displacement/enlongation (x)

Remember the elastic potential energy stored is equal to the area under the curve.

F = kx(N)

x (m)

F = kx(N)

x (m) 6

1 work unit

3 3

4 work units

elongation of 3 x units

will give one unit of

work done or energy stored.

stretching the spring TWICE as far to 6 x units

will require more work (and effort).

Doubling the stretch will require FOUR

times the amount of work done and energy

stored.EPE = 1/2 k x2

EPE is directly proportional to the square of the

elongation. Stretching the spring twice as far requires

twice the force but FOUR times WORK.

Dec 1512:25 PM

Resistance Bands and EPE

Resistance bands are used for resistance training! These bands allow us to get a 'workout' them because stretching the bands

requires AND expends energy.

Resistance bands are available in different tensions (spring constants) and are color coded accordingly.

Dec 1512:25 PM

Elastic Potential EnergyWork done when varying the spring constant (k).

EPE = 1/2 k x2

EPE is directly proportional to the value for the spring constant! Similar displacements require different amounts of work. The large spring constant requires more work and stores more elastic potential energy with similar elongation.

F(N)

x (m)

F(N)

x (m)

large spring constant

small area = small worklarge area = large work

small spri

ng constan

t

Dec 1512:31 PM

21 Determine the elastic potential energy stored in a spring whose spring constant is 250 N/m and which is compressed 8 cm.

Answ

er

https://www.njctl.org/video/?v=osLRYNZUXR8

Dec 1512:37 PM

22 What is the spring constant of a spring that is compressed 5 cm and has 0.65 J of elastic potential energy stored in it?

Answ

er




https://www.njctl.org/video/?v=GuIp7jaJ8E

https://www.njctl.org/video/?v=osLRYNZUXR8https://www.njctl.org/video/?v=GuIp7ja-J8E


13

November 04, 2016

Dec 1512:27 PM

A roller coaster is at the top of a track that is 80 m high. How fast will it be going at the bottom of the hill?

Eo + W = Ef

Eo = Ef

GPE = KE

mgh = 1/2mv2

v2 = 2gh

v2 = 2 (9.8m/s2) 80m v =39.6 m/s

W = 0

E0 = GPE, Ef = KE

Substitute GPE and KE equations

Solving for v yields


Mar 37:19 PM

23 A spring gun with a spring constant of 250 N/m is compressed 5 cm. How fast will a 0.025 kg dart move when it leaves the gun?

Answ

er

https://www.njctl.org/video/?v=qA9MABYv0uc

Mar 37:21 PM

24 A spring gun with a spring constant of 250 N/m is compressed 15 cm. How fast will a 0.025 kg dart go when it leaves the gun?

Answ

er

https://www.njctl.org/video/?v=tLxezl4ohfg

Mar 37:29 PM

25 A student uses a spring (with a spring constant of 180 N/m) to launch a marble vertically into the air. The mass of the marble is 0.004 kg and the spring is compressed 0.03m. What is the maximum height the marble will reach?

Answ

er

https://www.njctl.org/video/?v=uCMQdPMV7SA

Mar 37:36 PM

26 A student uses a spring gun (with a spring constant of 120 N/m) to launch a marble vertically into the air. The mass of the marble is 0.002 kg and the spring is compressed 0.04 m.How fast will the marble be traveling when it leaves the gun?

Answ

er

https://www.njctl.org/video/?v=12z8LAx_9no

Mar 37:43 PM

27 A roller coaster has a velocity of 25 m/s at the bottom of the first hill. How high was the hill?

Answ

er

https://www.njctl.org/video/?v=rN8dHTR4eL0

https://www.njctl.org/video/?v=qA9MABYv0uchttps://www.njctl.org/video/?v=tLxezl4ohfghttps://www.njctl.org/video/?v=uCMQdPMV7SAhttps://www.njctl.org/video/?v=12z8LAx_9nohttps://www.njctl.org/video/?v=rN8dHTR4eL0


14

November 04, 2016

Mar 37:51 PM

28 A student uses the lab apparatus shown above. A 5 kg block compresses a spring 6 cm. The spring constant is 300 N/m.What is the blocks velocity be when the spring loses all of the stored elastic potential energy?

Answ

er

https://www.njctl.org/video/?v=pCFbGfs3FFg

Mar 37:55 PM

29 How much work is done in stopping a 5 kg bowling ball rolling with a velocity of 10 m/s?

Answ

er

https://www.njctl.org/video/?v=jYIHIVoUs2M

Mar 37:58 PM

30 How much work is done in compressing a spring with a 450 N/m spring constant a distance of 2 cm?

Answ

er

https://www.njctl.org/video/?v=dTCGGwafpIc

Power


Power

https://www.njctl.org/video/?v=BRuTWJSCRu8

Jun 511:37 AM

Power

It is often important to know not only if there is enough energy available to perform a task but also how much time will be required.

Power is defined as the rate that work is done (or energy is transformed) :

W t

P =

100 Watt light bulbs convert 100 Joulesof electrical energy to heat and lightevery second.

Jun 511:37 AM

Power

Since work is measured in Joules (J) and time is measured in seconds (s) the unit of power is Joules per second (J/s).

However, in honor of James Watt, who made critical contributions in developing efficient steam engines, the unit of power is also know as a Watt (W).

W t

P =

https://www.njctl.org/video/?v=pCFbGfs3FFghttps://www.njctl.org/video/?v=jYIHIVoUs2Mhttps://www.njctl.org/video/?v=dTCGGwafpIchttps://www.njctl.org/video/?v=BRuTWJSCRu8


15

November 04, 2016

Jun 2812:28 AM

Since W = Fd parallel

Regrouping this becomes

Since v = d/t

Power

So power can be defined as the product of the force applied and the velocity of the object parallel to that force.

Jun 2812:33 AM

A third useful expression for power can be derived from our original statement of the conservation of energy principle.

Power

So the power absorbed by a system can be thought of as the rate at which the energy in the system is changing.

Since W = Ef E0

Jun 2812:33 AM

31 A steam engine does 50 J of work in 12 s. What is the power supplied by the engine?

Answ

er

https://www.njctl.org/video/?v=NXLuYIRc9F4

Jun 2812:35 AM

32 How long must a 350 W engine run in order to produce 720 kJ of work?

Answ

er

https://www.njctl.org/video/?v=IRYweBGBdI

Jun 2812:36 AM

33 A 12 kW motor runs a vehicle at a speed of 8 m/s. What is the force supplied by the engine?

Answ

er

https://www.njctl.org/video/?v=2O7vv19xrmo

Jun 2812:36 AM

34 An athlete pulls a sled with a force of 200N burning 600 Joules of food/caloric energy every second. What is the velocity of the athlete?

Answ

er

https://www.njctl.org/video/?v=TiRXw1qIltg

https://www.njctl.org/video/?v=NXLuYIRc9F4https://www.njctl.org/video/?v=IR-YweBGBdIhttps://www.njctl.org/video/?v=2O7vv19xrmohttps://www.njctl.org/video/?v=TiRXw1qIltg


16

November 04, 2016

Mar 38:30 PM

35 A 3.0 kg block is initially at rest on a frictionless, horizontal surface. The block is moved 8.0m in 2.0s by the application of a 12 N horizontal force, as shown in the diagram below. What is the power developed when moving the block?

A 24 B 32 C 48 D 96

8.0 m

3.0 kgF = 12 N

Frictionlesssurface

https://www.njctl.org/video/?v=GotMGnt9Idw

GPE, KE and EPE Problem

36 A spring launcher fires a marble at an angle of 520. In calculating the energy available for transformation into GPE and KE, what value of x is used in EPE = 1/2 kx2?

A The horizontal displacement of the spring.

B The vertical displacement of the spring.

C The straight line displacement of the spring, independent of the x and y axes.

Answ

er

C

Energy Problem Solving

Energy Problem SolvingWhat is the final velocity of a box of mass 5.0 kg that slides 6.0 m down a frictionless incline at an angle of 420 to the horizontal?

vo = 0

v = ?

The system will be the block. Since there is no friction, there are no external forces and we can use the Conservation of Energy (the gravitational force is taken care of by GPE). What types of energy are involved here?



Only three types of energy have been discussed so far. And in this case, there is only GPE and KE:

vo = 0

d=6.0m

θ=420

m = 5.0 kg

v = ?

(KE + EPE +GPE)0 = (KE + EPE + GPE)

becomes:

(KE + GPE)0 = (KE +GPE)

to save subscripts, we'll assume that no subscript implies a final quantity (KE = KEf)



vo = 0

d=6.0m

θ=420

m = 5.0 kg

v = ?

h0=4.0m

Let's put in the equations now:

Given (and using trig to find the value of h0 from d):

Find: v



vo = 0

d=6.0m

θ=420

m = 5.0 kg

v = ?

h0=4.0m

The velocity at the bottom of the incline is 8.9 m/s.

https://www.njctl.org/video/?v=GotMGnt9Idw


17

November 04, 2016



.

Consider the inclined plane problem that was just worked, but add a spring at the bottom of the incline.

The spring will be compressed a distance Δx and then released. Find the velocity of the box when it rises back to where it was first compressed a height of Δh.

What energies do we have to consider?



.

Once compressed, the box has EPE, GPE and zero KE.

When it loses touch with the spring at Δh above its fully compressed point, it will have KE, GPE and zero EPE.

(KE + EPE + GPE)0 = (KPE + EPE + GPE)

becomes:

(EPE + GPE)0 = (KE + GPE)



.

Before we proceed further with the solution, think how hard this problem would be to solve without using Conservation of Energy.

Once the object is released and the spring starts moving away from its compressed state, the force is no longer constant it will require mathematical integration (calculus) to solve. Free body diagrams are not the best way to find the velocity of the object.



.

Let's put in the equations and rearrange them to solve for v.



.

We now have the equation for the velocity when the block rises a vertical displacement of Δh.

But if we're only given Δx, how do we find Δh?

Use trigonometry and recognize that Δh = Δxsinθ.

Energy Problem

37 A box on an inclined plane is in contact with a spring. The box is released, compressing the spring. For every increment Δx, the box moves down the incline, how much does its height, Δh, change?

A Δx2

B Δxcosθ

C Δxsinθ

D √Δx

E Δxtanθ

Answ

er

C


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November 04, 2016

Energy Problem

38 A box is held on top of a spring on an inclined plane of angle θ = 310. The box is released, compressing the spring. If the spring moves 7.0 m down the plane, how much does its height, Δh, change?

A 7.0 m

B 6.5 m

C 6.0 m

D 3.6 m

E 3.0 m

Answ

er

D

Energy Problem

39 A box of mass m is on an inclined plane that makes an angle of θ with the horizontal and is in contact with a spring of spring constant k. The box is released, and it compresses the spring an amount Δx before rebounding. In terms of m, g, k and θ, what is the value of Δx?

Answ

erEnergy Problem

40 A spring (k = 150 N/m) on an incline of θ = 540 is compressed a distance of Δx = .060 m along the incline by a mass of 0.042 kg and then released. What is its velocity when it passes the point where it was first compressed and loses touch with the spring?

Answ

er

Energy Problem

41 A marble launcher shoots a marble vertically and then shoots a marble in the horizontal direction. Describe why the exit velocity of the marble in the two cases is different. Which exit velocity is greater?

Answ

er

In both cases, the EPE is the same. For the vertical case, part of the EPE is transformed into an increase in GPE as the exit point of the marble is higher than its compressed point. In the horizontal direction, there is no ΔGPE, hence the EPE is transformed totally into KE. The marble's velocity is greater in the x direction.

Conservation of Energy Problem Solving

Conservation of Energy Problem Solving


Falling Objects

Falling Objects the Energy way

.

An object, at rest, falls from a height, h0, to the ground, and you want to find out what its velocity is right before it hits the ground (assume no air friction).

Before you learned the Conservation of Energy, you would draw a free body diagram and then use a Kinematics equation to find the velocity. Review this with your group and then remove the screen below to check:

mg


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November 04, 2016

Falling Objects

Falling Objects the Energy way

Now, let's use the Conservation of Energy to solve this problem. Define the system as the object and GPE at the ground as zero. Since there is no friction, the work on the system will be zero.

In this case, both methods take about the same amount of effort. But, in cases where the motion, or the forces acting on the object are more complex or are changing, the Energy approach is easier.

Falling Objects problem

42 A ball of mass 0.45 kg falls from a building of height = 21 m. What is the ball's speed right before it hits the ground?

A 14 m/s

B 20 m/s

C 210 m/s

D 410 m/s

Answ

er

B

Falling Objects problem

43 Two objects, one with a mass of 0.43 kg, and the other with a mass of 42.5 kg, fall from a height of 31.1 m. Which object has the greater velocity right before it hits the ground? (assume no friction)

A Both have the same velocity.

B The 0.43 kg object

C The 42.5 kg object

Answ

er

A

Spring Problem

44 A spring gun, aimed in the horizontal direction with k = 250 N/m is compressed 0.05 m and released. How fast will a 0.025 kg dart go when it exits the gun?

Answ

er

Spring Problem

45 A student uses a spring, with k = 180 N/m, to launch a marble vertically into the air. The mass of the marble is 0.0040 kg and the spring is compressed 0.030 m. How high will the marble go above its initially compressed position?

Answ

er

Spring Problem

46 A student uses a spring gun (k = 120 N/m) to launch a marble at an angle of 520 to the horizontal (m = .0020 kg, Δx = 0.041 m). What is the maximum height that the marble will reach above its initially compressed position?

Answ

er


20

November 04, 2016

GPE Problem

47 A roller coaster car is pulled up to a height (A) of 50 m, where it then goes down the other side of the track. It traverses two other loops, one at a height of 40 m (B), and the second at a height of 30 m (C). Rank the velocities of the car at the three heights from greatest to least.

A A > B > C

B A > C > B

C B > A > C

D C > B > A

E C > B > A

F C > A > B

Answ

er

E

Energy Problem

48 Four objects are thrown with identical speeds in different directions from the top of a building. Which will be moving fastest when it strikes the ground? A

B

C

D

E All will have the same speed.

h Ans

wer

E

Energy Problem

49 Four objects are thrown with identical speeds in different directions from the top of a building. Which will hit the ground first?

A B

Answ

er

D

C

D

E All will hit at the same time.

h

Energy Problem

50 Four objects are thrown with identical speeds in different directions from the top of a building. Which will go the highest?

A B

Answ

er

A

C

D

E All will reach the same height.

h

Energy Problem

51 Four objects are thrown with identical speeds in different directions from the top of a building. Which will land furthest from the base of the building? A

B

Answ

er

B

C

D

E All will land at the same place.

h

Energy Problem

52 Four objects are thrown with identical speeds in different directions from the top of a building. Which will have the greatest horizontal component of its velocity at its maximum height?

A B

Answ

er

C

C

D

E All will be the same.

h


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November 04, 2016

Energy Problem

53 Three objects are thrown with identical speeds in different directions from the top of a building. Which will have the greatest kinetic energy at its maximum height?

A B

C

D All will have the same.

h Ans

wer

C

GPE and Escape Velocity

GPE and Escape Velocity


GPE expanded

GPE expanded

The expression GPE = mgh only works near the surface of and is specific to the planet. Recall that:

As an object's height above the surface of the planet increases, its distance from the center of the planet is not just the radius of the planet. Thus, a more general expression is needed based on Newton's Universal Gravitational Force equation:

Rplanet = radius of planetMplanet = mass of planet

rplanetobject = distance from center of the planet to the object

GPE expanded

GPE expandedDeriving a more general formula for the potential energy of an object that moves further away from the earth requires calculus, because the gravitational force is not constant it must be "integrated (calculus)" over the distance the object moves.

We'll just present the result, and use a more common physics term for gravitational potential energy, UG.

What do you think the negative sign in this expression means?

GPE expanded

GPE expanded

In solving the integral, an assumption was made that the gravitational potential energy between two objects an infinite distance apart is zero.

Add this assumption to the fact that the gravitational force is always an attractive force, and the negative sign pops up.

How can we reconcile this negative sign with the fact that we normally think of Gravitational Potential Energy as a positive number? Think of what GPE really means....

GPE expanded

GPE expanded

UG

GPE measured for objects near the surface of the earth is really mg(hh0), and h0 is normally set to zero.

So, it's the change in height above the earth's surface that is important GPE is not an absolute number.

On the graph above, notice as r (the distance from the center of the earth) increases (the object moves above the surface of the earth), UG becomes less negative. The difference between the final UG and the initial UG will be positive! So, mgh will still be a positive number.


22

November 04, 2016

GPE problem

54 Why can't the expression GPE = mgh be used to determine the GPE between the earth and a weather satellite in geosynchronous orbit above the earth (35.8 km)? Select two answers.

A The satellite's orbital period is twice that of the earth's rotational period.

B The satellite is moving at a great speed.

C The value of g is different at the earth's surface and at the satellite.

D The distances between the surface of the earth and the satellite from the center of the earth are mathematically significant.

Answ

er

C, D

GPE problem

55 Describe how the negative sign in the generalized potential energy equation was determined.

Answ

erEscape Velocity

Escape Velocity

Let's apply this new definition of GPE between two masses to the problem of escape velocity.

Escape velocity is the minimum velocity an object (such as a spaceship) needs to attain to "escape" the earth's gravitational pull. Actually, since the force of gravity has an infinite range, this is never really possible, but the force will be extremely tiny at great distances.

We're going to use conservation of energy, and our system is going to be the earthspaceship system. If we're looking for the minimum velocity, what will the magnitude of the potential energy and kinetic energy of the spaceship when it is an infinite distance from the earth?

Escape Velocity

Escape Velocity

Since we're looking for the minimum required velocity to escape the earth, we want it to get to infinity with zero kinetic energy. Potential energy was already defined to be zero at infinity. Using the Conservation of Energy:

Escape Velocity

Escape Velocity

Note that escape velocity is independent of the mass of the escaping object. It only depends on the mass and radius of the object being escaped from.

Using this fact, can you explain why despite the fact that Hydrogen and Helium are the most abundant elements in the universe, there are only trace amounts of each element in our atmosphere?

Escape Velocity

Escape Velocity

When the earth was first formed, the atmosphere contained significant amounts of both elements.

However Hydrogen and Helium are the lightest elements, so for a given atmospheric temperature, they would move faster than the heavier elements.

Over time, their average speeds exceeded the escape velocity of earth, thus they left the planet.


23

November 04, 2016

Escape Velocity Problem

56 A rocket of mass 5,000 kg will have ____________escape velocity than a rocket with mass 10,000 kg.

A a larger

B a smaller

C an equal

Answ

er

C

Escape Velocity Problem

57 What is the escape velocity on the planet Earth? (Mearth = 5.97x1024 kg, re = 6.38x106 m, G = 6.67x1011 Nm2/kg2)

Answ

erNov 49:09 AM Nov 49:05 AM

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