QA Assignment 14

#$&*

course PHY 121

7/19 3:40

014. Potential energy; conservative and non-conservative forces. *********************************************

Question: `q001. An automobile of mass 1500 kg coasts from rest through a displacement

of 200 meters down a 3% incline. How much work is done on the automobile by its weight

component parallel to the incline?

If no other forces act in the direction of motion (this assumes frictionless motion,

which is of course not realistic but we assume it anyway because this ideal situation

often gives us valuable insights which can then be modified to situations involving

friction), what will be the final velocity of the automobile?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

F = m * a

F = 1500 kg * 9.8 m/s^2

F = 14,700 N

3% incline: 14700 N * .03 = 441 N

Fnet = 14,700 N + 441 N

Fnet = 15,141 N

`dw = F & `ds

`ds = 15,141 N * 200 m

`dw = 3,028,200 J

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

The weight of the automobile is 1500 kg * 9.8 m/s^2 = 14,700 Newtons. The weight

component parallel to the incline is therefore very close to the small-slope

approximation

weight * slope = 14700 Newtons * .03 = 441 Newtons (small-slope approximation).

If no other forces act parallel to the incline then the net force will be just before

441 Newtons and the work done by the net force will be

`dWnet = 441 Newtons * 200 meters = 88200 Joules.

[ Note that this work was done by a component of the gravitational force, and that it

is the work done on the automobile by gravity. ]

The net work on a system is equal to its change in KE. Since the automobile started

from rest, the final KE will equal the change in KE and will therefore be 88200 Joules,

and the final velocity is found from 1/2 m vf^2 = KEf to be

vf = +_`sqrt(2 * KEf / m) = +-`sqrt(2 * 88200 Joules / (1500 kg) ) = +-`sqrt(2 * 88200

kg m^2/s^2 / (1500 kg) ) = +- 10.9 m/s (approx.).

Since the displacement down the ramp is regarded as positive and the automotive will end

up with a velocity in this direction, we choose the +10.9 m/s alternative.

STUDENT QUESTION: I assume the perpendicular force is balanced beause of the normal

force downward of gravity and mass and then upward from the road.

INSTRUCTOR RESPONSE:

Good, but there's a little more to it:

The normal force balances the component of the gravitational force which is

perpendicular to the road. The component of the gravitational force parallel to the

incline is the for that tends to accelerated objects downhill.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I got the force for the car and the force for the incline correct. I thought that I had

to put them together, but I can see why the slope is what you use to find the work for

this.

------------------------------------------------

Self-critique rating: 2

*********************************************

Question: `q002. If the automobile in the preceding problem is given an initial

velocity of 10.9 m/s up the ramp, then if the only force acting in the direction of

motion is the force of gravity down the incline, how much work must be done by the

gravitational force in order to stop the automobile?

How can this result be used, without invoking the equations of motion, to determine how

far the automobile travels up the incline before stopping?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

.5mvf^2 - .5mv0^2 = KE

.5(1500 kg)(0m/s)^2 - .5(1500 kg)(10.9m/s)^2 = KE

-89,107.5 = KE

It would take 89,107.5 Joules to stop the car.

KE = F * `ds

`ds = KE/F

`ds = 89,107.5 Joules/14700 N

`ds = 6.06 m

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

This is an application of the work-kinetic energy theorem.

In words, this theorem says that

the change in KE is equal to the work done by the net force acting ON the system

In symbols, this is expressed

`dW_net = `d(KE).

KE is kinetic energy, equal to 1/2 m v^2.

The automobile starts out with kinetic energy

KEinit = 1/2 m v0^2 = 1/2 (1500 kg) ( 10.9 m/s)^2 = 88000 kg m^2/s^2 = 88000 Joules.

The gravitational force component parallel to the incline is in this case opposite to

the direction of motion so that gravity does negative work on the automobile. Since the

change in KE is equal to the work done by the net force acting ON the system, if the

gravitational force component parallel to the incline does negative work the KE of the

object will decrease. This will continue until the object reaches zero KE.

As found previously the gravitational force component along the incline has magnitude

441 Newtons. In this case the forces directed opposite to the direction motion, so if

the direction up the incline is taken to be positive this force component must be -441

N. By the assumptions of the problem this is the net force exerted on the object.

Acting through displacement `ds this force will therefore do work `dWnet = Fnet * `ds =

-441 N * `ds. Since this force must reduce the KE from 88,000 Joules to 0, `dWnet must

be -88,000 Joules. Thus

`dW_net = -441 N * `ds = -88,000 Joules

and

`ds = -88,000 J / (-441 N) = 200 meters (approx.).

Had the arithmetic been done precisely, using the precise final velocity found in the

previous exercise instead of the 10.9 m/s approximation, we would have found that the

displacement is exactly 200 meters.

STUDENT QUESTION

I set `dKE = Fnet * `ds and had `dKE as negative so I came up with the correct solution

but just above you say that work is negative.

I don’t understand how work is negative, especially going by the equation because I

thought work was opposite `dKE.

INSTRUCTOR RESPONSE

You're thinking about exactly the right things.

The specific statement of the work-KE theorem is that the work done by the net force

acting ON the system is equal to the change in the kinetic energy of the system. This is

abbreviated

`dW_ON_net = `dKE.

Rather that talking about 'the work', it's very important to get into the habit of

labeling 'the work' very specifically. You have two basic choices. You can think in

terms of

the work done on the system or object by a force

the work done by the system or object against a force

The two are equal and opposite.

Note that the words 'on' and 'by' modify the word 'system', not the word 'force'.

The key phrases are 'on the system' and 'by the system'.

In the present case if you were to choose to think in terms of the work done by the net

force exerted by object, then this force would be labeled `dW_BY_net and would be equal

and opposite to `dW_ON_net, the work exerted by the net force acting on the object.

Formally we would have

`dW_BY_net = - `dW_ON_net so that

`dW_BY_net = - `dKE.

This last equation is often written

`dW_BY_net + `dKE = 0,

and is another equivalent formulation of the work-kinetic energy theorem.

STUDENT COMMENT

I am getting all the equations mixed up is there any way you can just send the different

equations? I understand the 4 from the major quiz. I can do the algebra I just don’t

know which equation to plug it in for.

INSTRUCTOR RESPONSE

Physics is about more than figuring out what to plug into what equation. It's necessary

to understand the words and the concepts to know which equation to plug into. In other

words, the concepts are what keep us from getting the equations mixed up.

However I have observed in your work that you do very well with the algebra. So the

equations might well be your most appropriate starting point. You can use the equations

to understand the words and the concepts, just as less algebraically adept students

might use the words and concepts to understand the equations.

The relevant relationships here are

`dW_net = `dKE and

KE = 1/2 m v^2.

The relationship

`dW_BY_net = - `dW_ON_net

is also invoked in the additional comments at the end, which mention an alternative

formulation of the work-kinetic energy theorem. However this relationship is not used in

solving this particular problem.

STUDENT COMMENT

OK, I understand the solution and will use _on and _by descriptors in my

answers from now on.

INSTRUCTOR RESPONSE

Good.

Remember that ON and BY are adjectives applied to the word 'system', not to the word

'force'.

That is, you have to determine whether the force is acting ON the system, or is exerted

BY the system.

Your choice of point of view will determine whether you use the equation

`dW_NC_ON = `dPE + `dKE

or

`dW_NC_BY + `dPE + `dKE = 0.

STUDENT COMMENT

Ok. I see why my ‘ds was negative. The F is negative in this system because it is

working against the positive motion of the car UP the ramp. For every force, there is an

equal and OPPOSITE force.

INSTRUCTOR RESPONSE

Good. If you assume the positive direction to be up the incline, F_net is negative, as

you say, but the specific reason is slightly different than the one you give. It's good

to think in terms of equal and opposite forces, but the motion of the car is not a

force.

In this case it really just comes down to signs:

The force used to calculate `dW_net was the net force acting on the car. That force acts

down the incline, in the direction opposite motion. Therefore F_net and `ds are of

opposite sign, and the net force acting on the car does negative work. This decreases

the KE, as your solution indicates.

The question of whether `ds is positive or negative depends on which direction you

choose for the positive direction. In your solution you apparently thought of upward as

the positive direction; you should have explicitly stated this. Relative to this choice

F_net is negative.

As I mentioned, you should have declared the positive direction in your solution. You

could have chosen either upward (which is the direction of the displacement, and is the

direction you implicity chose) or downward (which is the direction of the net force) to

be the positive direction.

Either choice of positive direction would have been perfectly natural. If you had chosen

'down the incline' to be the positive direction, then `ds would have been negative

(therefore opposite to the downward direction, so up the incline). In either case, `ds

would have been up the incline.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

If I had used the v0^2 from the prior problem, I would have gotten 88,200 rather than

89,107.5.

I do understand where I went wrong with the problem, but I do not promise that I have

internalized it to the point that I won't go wrong again.

------------------------------------------------

Self-critique rating: 2

@&

I believe those two results differ only by a roundoff error in one of the quantities.

*@

*********************************************

Question: `q003. If the automobile in the previous example rolls from its maximum

displacement back to its original position, without the intervention of any forces in

the direction of motion other than the parallel component of the gravitational force,

how much of its original 88200 Joules of KE will it have when it again returns to this

position?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

F = 1500 kg * 9.8 m/s^2

F = 14,700 N

14,700 N * .03 = 441 N

`dKE = F * `ds

`dKE = 441N * 200 m

`dKE = 88,200 Joules

It should get all of the original energy back.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

In the first exercise in the present series of problems related to this ramp, we found

that when the automobile coasts 200 meters down the incline its KE increases by an

amount equal to the work done on it by the gravitational force, or by 88200 Joules.

Thus the automobile will regain all of its 88200 Joules of kinetic energy.

To summarize the situation here, if the automobile is given a kinetic energy of 88200

Joules at the bottom of the ramp then if it coasts up the ramp it will coast until

gravity has done -88200 Joules of work on it, leaving it with 0 KE. Coasting back down

the ramp, gravity works in the direction of motion and therefore does +88200 Joules of

work on it, thereby increasing its KE back to its original 88200 Joules.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating: 4

*********************************************

Question: `q004. Explain why, in the absence of friction or other forces other than

the gravitational component parallel to incline, whenever an object is given a kinetic

energy in the form of a velocity up the incline, and is then allowed to coast to its

maximum displacement up the incline before coasting back down, that object will return

to its original position with the same KE it previously had at this position.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

An object, based on gravity, the incline, and its mass has the potential to move a

certain distance. In the case of coasting up the hill, it uses all of this energy and

comes to a stop. Based on those same factors of gravity, incline, and mass, the car

still has that much potential energy at that point and will go back down to where it

started and the energy will go back to zero.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

The car initially had some KE. The gravitational component parallel to the incline is

in the direction opposite to the direction of motion up the incline and therefore does

negative work ON the object as it travels up the incline.

The gravitational component is the net force on the object, so the work done by this net

force on the system causes a negative change in KE, which eventually decreases the KE to

zero so that the object stops for an instant. This happens at the position where the

work done BY the net force is equal to the negative of the original KE.

The gravitational component parallel to the incline immediately causes the object to

begin accelerating down the incline, so that now the parallel gravitational component is

in the same direction as the motion and does positive work ON the system.

At any position on the incline, the negative work done by the gravitational component as

the object traveled up the incline from that point, and the positive work done by this

force as the object returns back down the incline, must be equal and opposite.

This is because the displacement up the incline and the displacement down the incline

are equal and opposite, while the parallel gravitational force component remains the

same. Thus the Fnet * `ds products for the motion up and the motion down equal and

opposite.

When the object reaches its original point, the work that was done on it by the net

force, as it rolled up the incline, must be equal and opposite to the work done on it

while coasting down the incline. Since the work done on the object while coasting up

the incline was the negative of the original KE, the work done while coasting down,

being the negative of this quantity, must be equal to the original KE. Thus the KE must

return to its original value.

STUDENT QUESTION

I still really don’t understand how it can return back to its original position because

of what we saw

in class It never returned back to its original position.

INSTRUCTOR RESPONSE

There are a number of situations in which an object doesn't return to its original

position.

The one that's relevant to this situation:

When you rolled the ball up the single incline, it slowed, came to rest for an instant,

and then rolled back down. It did return to its initial position. Of course when it got

there is was moving pretty fast so if you didn't stop it, it kept going until something

else did.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating: 3

*********************************************

Question: `q005. As the object travels up the incline, does gravity do positive or

negative work on it? Answer the same question for the case when the object travels down

the incline.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

As gravity travels up an incline, gravity does negative work on it; gravity is pulling

it back. When an object travels down an incline, gravity does positive work on it;

gravity helps it along.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

As the object travels up the incline the net force is directed opposite its direction of

motion, so that Fnet_ON and `ds have opposite signs and as a result `dWnet = Fnet_ON *

`ds must be negative.

As the object travels down the incline the net force is in the direction of its motion

so that Fnet_ON and `ds have identical signs and is a result `dWnet = Fnet_ON * `ds must

be positive.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating: 3

*********************************************

Question: `q006. If positive work is done on the object by gravity, will it increase

or decrease kinetic energy of the object?

Answer the same question if negative work is done on the object by gravity.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Positive work done by gravity on an object will increase the kinetic energy of the

object. If negative work is done by gravity on the object, the kinetic energy will

decrease.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

The KE change of an object must be equal to the work done ON the system by the net

force. Therefore if positive work is done on an object by the net force its KE must

increase, and if negative work is done by the net force the KE must decrease.

STUDENT QUESTION (instructor comments in parentheses)

Ok, after reading a couple of time I get it dw will be equal to KE. If dw is - Ke will

be - .

KE can't be negative; `dW is not equal to KE, but to `dKE, the change in KE.

`dKE can certainly be positive or negative (or zero), depending on the situation.

I am still a little unclear about if the dw done on an object is negative then what

direction is it

moving??

The sign of `dW by the net force does not determine the direction of motion of the

object. It determines only the change in its kinetic energy.

In the present case, the net force is the component of gravity along the incline. The

direction of motion of the object determines whether this force is in the direction of

motion or opposite that direction, and so determines whether the displacement is in the

direction of motion (implying positive work) or opposite the direction of motion

(implying negative work).

The direction of motion thus determines, for this situation, whether `dW_net is positive

or negative.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating: 3

*********************************************

Question: `q007. While traveling up the incline, does the object do positive or

negative work against gravity?

Answer the same question for motion down the incline.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

If you are assuming that up the incline is the positive direction, thene th object is

doing positive work agains gravity. If down it the incline is positive, then the object

is moving in the direction that the gravity is pulling it. When gravity is positive,

then the work by the object is negative.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

If the object ends up in the same position as it began, the work done on the object by

gravity and work done by the object against gravity must be equal and opposite. Thus

when the object does positive work against gravity, as when it travels up the incline,

gravity is doing negative work against the object, which therefore tends to lose kinetic

energy.

When the object does negative work against gravity, as when traveling down the incline,

gravity is doing positive work against the object, which therefore tends to gain kinetic

energy.

STUDENT COMMENT:

A little shaky on this problem because I feel its easy to get confused on the positive

and negative.

INSTRUCTOR RESPONSE

This is the most common point of confusion at this stage of the course.

To sort out positive and negative, you would answer the following questions:

Are you thinking about the work done ON the system or BY the system (i.e., are you

thinking about the forces acting ON the system or a forces exerted BY the system)? The

ON and the BY are equal and opposite.

Whichever force you are thinking about, it does positive work when it is in the

direction of motion and negative work when it is opposite the direction of motion.

STUDENT QUESTION

Ok, so i get his really mixed up. The work done BY the object is positve, against

gravity which is doing negative work ON

the object going up the inlcine. When going down the incline work done BY the object is

negative as work done ON the object by gravity is positive.

Is this right?

INSTRUCTOR RESPONSE

Your statement is correct.

And, until it 'clicks', this is certainly confusing. It takes most students a few

assignments before this becomes clear. You are progressing nicely.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): ok

------------------------------------------------

Self-critique rating: 2

*********************************************

Question: `q008. Suppose that the gravitational force component exerted parallel to a

certain incline on an automobile is 400 Newtons and that the frictional force on the

incline is 100 Newtons. The automobile is given an initial KE of 10,000 Joules up the

incline. How far does the automobile coast up the incline before starting to coast back

down, and how much KE does it have when it returns to its starting point?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Friction and gravity will both work on the automobile, so the forces are 400 Newtons +

100 Newtons = 500 Newtons. Because the work is done on the automobile, that will be a

negative number.

The initial kinetic energy is 10,000 Joules up the hill

`dw_net = fNet_on * `ds

`ds = `dw_net/fnet

`ds = -10,000 Joules/-500 N

`ds = 20 meters

KE = 500 N * 20 m

KE = 10,000 Joules

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

The net force ON the automobile as it climbs the incline is the sum of the 400 Newton

parallel component of the gravitational force, which is exerted down the incline, and

the 100 Newton frictional force, which while the automobile is moving up the incline is

also exerted down the incline. Thus the net force is 500 Newtons down the incline.

This force will be in the direction opposite to the displacement of the automobile up

the incline, and will therefore result in negative work being done on the automobile.

When the work done by this force is equal to -10,000 Joules (the negative of the

original KE) the automobile will stop for an instant before beginning to coast back down

the incline. If we take the upward direction to be positive the 500 Newton force must

be negative, so we see that

-500 Newtons * `ds = -10,000 Joules

so

`ds = -10,000 Joules / -500 Newtons = 20 Newtons meters / Newtons = 20 meters.

After coasting 20 meters up the incline, the automobile will have lost its original

10,000 Joules of kinetic energy and will for an instant be at rest.

The automobile will then coast 20 meters back down the incline, this time with a 400

Newton parallel gravitational component in its direction of motion and a 100 Newton

frictional force resisting, and therefore in the direction opposite to, its motion. The

net force ON the automobile will thus be 300 Newtons down the incline.

The work done by the 300 Newton force acting parallel to the 20 m downward displacement

will be 300 Newtons * 20 meters = 6,000 Joules. This is 4,000 Joules less than the when

the car started.

This 4,000 Joules is the work done during the entire 40-meter round trip against a force

of 100 Newtons which every instant was opposed to the direction of motion (100 Newtons *

40 meters = 4,000 Newtons). As the car coasted up the hill the frictional force was

downhill and while the car coasted down friction was acting in the upward direction.

Had there been no force other than the parallel gravitational component, there would

have been no friction or other nongravitational force and the KE on return would have

been 10,000 Joules.

STUDENT QUESTION

I’m still not sure how I would find the final KE with all the other forces. I read the

given

solution multiple times but I am still really confused.

INSTRUCTOR RESPONSE

Your solution was fine for motion up the incline, and agreed with the given solution to

that point.

The first line of the given solution that you didn't address in your solution begins

'The automobile will then coast 20 meters back down the incline, ... '

The point you need to understand is that the 400 N gravitational component is still

there, but since the car is moving down the incline the 100 N frictional force is now

directed up the incline. So the net force on the car is 300 N down the incline.

You therefore know the net force acting on the system, and you know the displacement, so

you can easily calculate the work `dW_ON done by this net force. This is equal to the

change in KE as the car travels down the ramp.

The car began with 0 KE at the top (it came to rest for an instant), so you can figure

out its final KE.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I did okay up to the coming back down the hill. It does make sense that gravity will

now work with the car while friction continues to work against it, so that net force is

400 N - 100 N = 300 N. When you multiply the 300 N times the 20 meters, you wll not

have the same amount of work. It will be 6,000 Joules.

------------------------------------------------

Self-critique rating: 2

*********************************************

Question: `q009. The 1500 kg automobile in the original problem of this section

coasted 200 meters, starting from rest, down a 3% incline. Thus its vertical

displacement was approximately 3% of 200 meters, or 6 meters, in the downward direction.

Recall from the first few problems in this q_a_ that the parallel component of the

gravitational force did 88200 Joules of work on the automobile. This, in the absence of

other forces in the direction of motion, constituted the work done by the net force and

therefore gave the automobile a final KE of 88200 Joules.

How much KE would be automobile gain if it was dropped 6 meters, falling freely through

this displacement?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

1500 kg * 9.8 m/s^2 = F

F = 14,700 Newtons

14,700 Newtons * 6 m = 88,200 Joules

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

Gravity exerts a force of

gravitational force = 1500 kg * 9.8 m/s^2 = 14700 N

on the automobile. This force acting parallel to the 6 meter displacement would do

`dW by gravitational force = 14700 N * 6 meters = 88200 Joules of work on the

automobile.

This is the work done by the entire gravitational force, not just by its component

parallel to the incline. It is multiplied by the 6 meter vertical displacement of the

automobile, since it acts along the same line as that displacement.

However this is identical to the work done on the automobile by the parallel component

of the gravitational force in the original problem. In that problem the parallel

component was multiplied by the displacement along the incline (which was much more than

6 meters), since it acts along the same line as that displacement.

STUDENT QUESTION

I do not see how it’s the same amount of work? 88200 Joules

INSTRUCTOR RESPONSE

The first few problems in this qa obtained KE = 88200 Joules.

They also obtained the result that the work done by the parallel component of the

gravitational force acting on the system was 88200 Joules.

In this problem we see that in a straight 6 meter drop the work done by gravity is the

same, 88200 Joules.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating: 3

*********************************************

Question: `q010. When the automobile was 200 meters 'up the incline' from the lower

end of the 3% incline, it was in a position to gain 88200 Joules of energy. An

automobile positioned in such a way that it may fall freely through a distance of 6

meters, is also in a position to gain 88200 Joules of energy. The 6 meters is the

difference in vertical position between start and finish for both cases.

How might we therefore be justified in a conjecture that the work done on an object by

gravity between two points depends only on the difference in the vertical position

between those two points?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The force of gravity remains the same at 9.8 m/s^2. The mass of the car also remains

the same. When the car traveled 200 m down a 3% incline, it was actually a vertical

drop of 6 meters. The same amount of work is done because that's the amount needed for

that vertical drop, whether it's straight down or over an incline.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

This is only one example, but there is no reason to expect that the conclusion would be

different for any other small incline. Whether the same would be true for a non-

constant incline or for more complex situations would require some more proof. However,

it can in fact be proved that gravitational forces do in fact have the property that

only change in altitude (or a change in distance from the source, which in this example

amounts to the same thing) affects the work done by the gravitational force between

points.

STUDENT QUESTION

I am not sure about this concept could you elaborate? I am not really even sure what the

question is asking in the first

place.

INSTRUCTOR RESPONSE

Subsequent questions will also reinforce this idea.

It was shown in previous problems that an automobile at the top of the ramp is in a

position to gain about 88000 J

of kinetic energy, by rolling without friction down the incline.

It was also shown that an automobile that falls freely from the top of the ramp to the

level of the bottom of the ramp will gain the same amount of kinetic energy.

In both cases the vertical position of the automobile changed by the same amount.

We therefore conjecture that there's something in the change in the vertical position of

the automobile that determines

how much energy it can gain or lose to gravity.

STUDENT QUESTION

I got it right. Still cant completely figure ou t the concept though.

------------------------------------------------

INSTRUCTOR RESPONSE

14 700 N raised 6 meters requires 88 200 Joules.

If you roll the car up a 3% incline you only have to exert 3% of the 14 700 N weight

(provided friction is negligible), but you have to travel 200 meters in order to raise

the car 6 meters. If you multiply force by displacement, you again get 88 200 Joules.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating: 3

*********************************************

Question: `q011. If an object has a way to move from a higher altitude to a lower

altitude then it has the potential to increase its kinetic energy as gravity does

positive work on it. We therefore say that such an object has a certain amount of

potential energy at the higher altitude, relative to the lower altitude. As the object

descends, this potential energy decreases. If no nongravitational force opposes the

motion, there will be a kinetic energy increase, and the amount of this increase will be

the same as the potential energy decrease. The potential energy at the higher position

relative to the lower position will be equal to the work that would be done by gravity

as the object dropped directly from the higher altitude to the lower.

A person holds a 7-kg bowling ball at the top of a 90-meter tower, and drops the ball to

a friend halfway down the tower. What is the potential energy of the ball at the top of

the tower relative to the person to whom it will be dropped?

How much kinetic energy will a bowling ball have when it reaches the person at the lower

position, assuming that no force acts to oppose the effect of gravity?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

F = 7 kg * 9.8 m/s^2

F = 68.6 Newtons

'dKE = F * `ds

`KE = 68.6 Newtons * 90 m

`KE = 6174 Joules to get to the bottom

It will still have half at the middle, so 1/2 of 6174 Joules = 3087 Joules

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

The potential energy of the ball at the top of the tower is equal to the work that would

be done by gravity on the ball in dropping from the 90-meter altitude at the top to the

45-meter altitude at the middle of the tower. This work is equal to product of the

downward force exerted by gravity on the ball, which is 7 kg * 9.8 m/s^2 = 68.6 Newtons,

and the 45-meter downward displacement of the ball. Both force and displacement are in

the same direction so the work would be

work done by gravity if ball dropped 45 meters: 68.6 Newtons * 45 meters = 3100 Joules,

approx..

Thus the potential energy of the ball at the top of the tower, relative to the position

halfway down the tower, is 3100 Joules. The ball loses 3100 Joules of PE as it drops.

If no force acts to oppose the effect of gravity, then the net force is the

gravitational force and the 3100 Joule loss of PE will imply a 3100 Joule gain in KE.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I guess I was thinking that it had the potential to get to the ground, it just wouldn't

use it all, but I understand the answer.

------------------------------------------------

Self-critique rating: 3

*********************************************

Question: `q012. If a force such as air resistance acts to oppose the gravitational

force, does this have an effect on the change in potential energy between the two

points? Would this force therefore have an effect on the kinetic energy gained by the

ball during its descent?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Air resistance would mean less motion/velocity. This would decrease the Kinetic

Energy. A decrease in Kinetic energy would mean in increase in potential energy.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

The change in potential energy is determined by the work done on the object by gravity,

and is not affected by the presence or absence of any other force. However the change

in the kinetic energy of the ball depends on the net force exerted, which does in fact

depend on whether nongravitational forces in the direction of motion are present.

We can think of the situation as follows: The object loses gravitational potential

energy, which in the absence of nongravitational forces will result in a gain in kinetic

energy equal in magnitude to the loss of potential energy. If however nongravitational

forces oppose the motion, they do negative work on the object, reducing the gain in

kinetic energy by an amount equal to this negative work.

ADDITIONAL INSTRUCTOR COMMENT:

The PE change is the result of only conservative forces; in this case the PE change

depends only on the vertical displacement and the gravitational force.

The force of air resistance is nonconservative and has no effect on PE change.

KE change depends on net force, which includes both conservative and nonconservative

forces. So KE change is affected by both PE change and the work `dW_nc_ON done by

nonconservative forces.

In this case PE change is negative, which tends to increase KE, while nonconservative

forces do negative work on the mass, which tends to reduce KE.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating: 2

*********************************************

Question: `q014. If an average force of 10 Newtons, resulting from air resistance,

acts on the bowling ball dropped from the tower, what will be the kinetic energy of the

bowling ball when it reaches the halfway point?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Fnet = F + F_air

Fnet = 68.6 N + (-10 N)

Fnet = 58.6 N

`dKE = 58.6N * 45 m

`dKE = 2637 Joules

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

The ball still loses 3100 Joules of potential energy, which in the absence of

nongravitational forces would increase its KE by 3100 Joules. However the 10 Newton

resisting force does negative work -10 N * 45 m = -450 Joules on the object. The object

therefore ends up with KE 3100 J - 450 J = 2650 J instead of the 3100 J it would have in

the absence of a resisting force.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I did the subtracting first and then did the multiplication. I got the same answer, but

am not sure if the thinking will always work out correctly.

@&

There are two equivalent ways to think of this, and yours was fine.

*@

------------------------------------------------

Self-critique rating: 2

If you understand the assignment and were able to solve the previously given problems

from your worksheets, you should be able to complete most of the following problems

quickly and easily. If you experience difficulty with some of these problems, you will

be given notes and we will work to resolve difficulties.

*********************************************

Question: `q015. An Atwood machine has a 5-kg mass on one side and a 6-kg mass on the

other. If the 6 kg mass descends 2 meters:

Was its displacement in the same direction as the gravitational force exerted on it, or

in the opposite direction?

What was the work done on it by gravity?

Was the displacement of the 5-kg mass in the same direction as the gravitational force

exerted on it, or in the opposite direction?

What was the work done by gravity on this mass?

Did gravity do net positive or negative work on this system?

What was the change in the gravitational PE of each object, and of the system?

Assuming no frictional or other dissipative forces, what was the change in the KE of the

system?

If the system started from rest, what was its velocity after having descended the 2

meters?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The displacement of the 6 kg mass is in the same direction as the gravitational force.

The work done by gravity on this mass is:

F=6 kg * 9.8 m/s^2

F = 58.8 N

`dw = F * `ds

`dw = 58.8 N * 2 m

`dw = 117.6 Joules

The displacement of the 5 kg mass is in the opposite direction as the gravitational

force exerted on it.

The work done by gravity on this mass is:

F = 5 kg * 9.8 m/s^2

F = 49 Newtons

`dw = 49 N * -2 m

`dw = -98 Joules

The net work done by gravity on this system is positive.

The change in the gravitational PE is equal to and opposite of the kinetic energy, so

the 6 kg mass is -117.6 Joules, and the 5 kg mass is 98 Joules.

The change of the kinetic energy of the system is 117.6 J - 98.0 J = 19.6 Joules.

Velocity would be KE= 1/2 mv^2

1/2 mv^2 = KE

mv^2 = 2KE

v^2 = 2KE/m

v = `sqrt (2KE/m)

v = `sqrt [(2 * 19.6 J)/11 kg]

v = `sqrt (39.2 J / 11 kg)

v = 3.6 m/s

@&

Very good.

*@

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!

@&

Well done. Check my notes.

*@