If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution.
If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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?
Your solution:
Confidence rating:
Given Solution:
The weight of the automobile is 1500 kg * 9.8 m/s^2 = 14,700
weight * slope = 14700
If no other forces act parallel to the incline then the net
force will be just before 441
`dWnet = 441
[ 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):
Self-critique rating:
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?
Your solution:
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
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):
Self-critique rating:
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?
Your solution:
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):
Self-critique rating:
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.
Your solution:
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):
Self-critique rating:
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.
Your solution:
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):
Self-critique rating:
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.
Your solution:
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):
Self-critique rating:
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.
Your solution:
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):
Self-critique rating:
Question:
`q008. Suppose that the gravitational force
component exerted parallel to a certain incline on an automobile is 400
Your solution:
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
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
-500
so
`ds = -10,000 Joules / -500
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
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
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):
Self-critique rating:
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?
Your solution:
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):
Self-critique rating:
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?
Your solution:
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):
Self-critique rating:
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?
Your solution:
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
work done by gravity if ball dropped 45 meters: 68.6
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):
Self-critique rating:
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?
Your solution:
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):
Self-critique rating:
Question:
`q014. If an average force of 10
Your solution:
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
Self-critique (if necessary):
Self-critique rating:
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?
Your solution:
Self-critique (if necessary):
Self-critique rating: