# Acceleration Force And Mass Essay Research Paper

Acceleration, Force And Mass Essay, Research Paper

Aim: ?I intend to

investigate the rule of F = M*A and so investigate the relationships

between acceleration, force and mass and how they affect each other.Preliminary Workings: The main aim of my project is to investigate three factors,

and so I will start off with a few lines about each of them. The idea of a force is

fundamental to physics and it is simply thought of as a push or a pull, but

this is not satisfactory for my purposes. We cannot see a force but instead we

can see its effect on an object (the principle of Brownian Motion), so forces

are described in terms of what they do. Forces tend to cause changes in an

objects 1.

Shape or size 2.

Speed in a straight line 3.

Direction Forces are measured in newtons (N), named after the person

who first invented this unit. When several forces act on an object, they can

either combine to give an overall force ? which will change the object?s shape

or motion ? or they could cancel each other out, giving no overall force. In

the last case it could be said that the forces are ?balanced?. If there is no

force acting, or if all the forces acting on an object are balanced, then there

will be no change taking place. An object at rest will remain at rest, and a

moving object will continue to move, keeping the same speed and travelling in

the same direction. ??????????? The mass

of an object tells us how much matter it contains and is measured in the unit

of kilograms (kg). Whereas mass is a scalar quantity (magnitude only), forces

are vector quantities, meaning they have both direction and magnitude. ??????????? Acceleration

is the rate at which the velocity of an object changes, over a period of

time. It is measured in metres-per-second per second (m/s/s) or

meters-per-second squared (m/s²), and it tells you how much the velocity will

change each second. The acceleration of an object can be calculated by using

the following formula:(average) acceleration (m/s²) =?? change in velocity (m/s)??????? or in symbols: a = v – u ??????????????????????????????????????????????? ? time taken for the change (s)?? ??????????????????????????????????? twhere u is the velocity at the beginning of the time

interval and v is the velocity at the end of the time interval. When an

object is slowing down the change in the velocity is negative (because v is

less than u), and so the acceleration is negative. This is sometime

called deceleration. The acceleration at any point on a? journey can be calculated by measuring the

slope of a velocity-time graph. In effect this is the same as applying the

formula that I have included above. To show some of these graphs I have

included some below to show how different accelerations can be portrayed for

varying and constant changes.??????????????????????????????????? Velocity (m/s)?????????????????????????????????????????? Velocity (m/s) ??????????? ?????????????????????????????????????????????????? Time (s)???????????????????????????????????????????????????????? Time (s) ??????? Steady

acceleration from rest???????????????????

Increasing acceleration from rest From these examples it is possible to work out what other

graphs of this type would look like and should stand me in good stead for this

project.The next scientific that I will look at is that worked on by

the famous scientist that I have already mentioned ? Sir Isaac Newton. During

his work he discovered some laws of motion, which are quite appropriate to what

I am investigating: Law 1: Any object will continue to do what it is

already doing unless a resultant force is acting on it.?????? I am used to the idea that an object on the ground, which is

given to start it, will come to rest quickly. Of course once it is moving,

friction is a force that acts upon it to cause a change, in this case a

reduction in velocity until the object stops. Without friction, as in space, an

object given a push will continue to move in a straight line with the velocity

it had at the end of the push. This can be showed using an air track or some

other method of reducing friction. Though the law refers to a resultant force.

So the other way in which an object can remain in a constant state is if the

resultant force acting on it is zero. Law 2: Constant acceleration causes

constant acceleration. The greater the force, the greater the acceleration for

that particular body. Therefore force is proportional to acceleration.? F??????

a If a particular acceleration is to be achieved, the force

required to achieve it is also dependant on the mass to be moved. So? F??????

m??????????? ??????????? Then F = ma If we define the unit of force such that 1 unit of force

will accelerate 1kg by 1 m/s² we have the definition of the Newton. This can

also be thought of by another means. The formula can be arranged to read that m

= F/a so that the bigger the mass the less the acceleration that could be

produced. One-way of thinking about mass is to regard it as the ?lack of

willingness to move? of an object. This property is sometimes called inertia.

The F in the equation is the resultant force acting on an object.Law 3: When an object is acted on by a force, then

somewhere another object is acted on by an equal force in the opposite

direction. This research has made me think about exactly how I am to

carry out this experiment. I feel that I have researched enough evidence but

now it is time for me to consider the method from which I will take my results

and with that my conclusions and evaluations. Firstly I will need to measure

the acceleration of an object and there is an instrument available to me that

can perform this task very well indeed. That is a ticker-timer. A piece of

ticker-timer tape is passed through the instrument and as an object moves along

(with one end of the tape attached) it is pulled through the ticker-timer. A

vibrating beam marks points on a piece of carbon paper and this in turn marks

along the tape and it does this at a rate of 200 dots per second. By taking a

sequence of these dots (for this experiment I will use 10) the acceleration of

the object can be determined. Therefore if the dots are equally spaced then the

acceleration is constant and if the spacing between the dots increases each

time, the acceleration is becoming greater. The next step is to use the formula

distance = speed * time to work out the velocity of the first ten dots

(from rest). The velocity is then taken for the next ten dots and the

acceleration between these two times is calculated. The formula for

acceleration, as previously stated is:(average) acceleration (m/s²) =?? change in velocity (m/s)??????? or in symbols: a = v – u ??????????????????????????????????????????????? ? time taken for the change (s)The time taken in this equation is that for ten dots and

this equates to a time period of 0.2 seconds.Preliminary Experiment and Method: ? Before any experiment is taken out, there should always be

an experimental trial to determine factors such as ranges and to make

alterations to the overall set-up to gather the most accurate data that is

possible. Firstly I have already stated what I hope to achieve, but little in

the way that it will actually be carried out. The formula of F = ma will

be applied to a wooden trolley, which will have its mass recorded before the

experiment, and have its acceleration measured throughout. A range of masses

(acting as the force for my purposes) will be attached to the front end of the

trolley by way of a pulley system (shown in my diagram) and attached to the other

end will be a piece of ticker-timer tape on which will generally record the

acceleration. The ticker-timer will be turned on and then the force will be

allowed to pull the trolley from its rest position while its acceleration is

recorded by way of the dots marked on the ticker-timer tape. When the trolley

has reached the end of its course the results will then be recorded. As I have

previously stated a preliminary test was done before final readings were taken

and here are the results: Mass /kg

(trolley) Force /N (force applied) Distance1 /m Velocity1 /m/s Distance2 /m Velocity2 /m/s Acceleration /m/s² 0.302 0.5 0.018 0.090 0.032 0.160 0.350 0.302 1.0 0.031 0.150 0.066 0.330 0.900 0.302 2.5 0.063 0.315 0.147 0.735 2.100 0.302 3.0 0.077 0.385 0.173 0.865 2.400 0.302 4.5 0.093 0.465 0.215 1.075 3.050 ? I was pleased with what I managed to take out from my

preliminary experiment. The ranges used for the results above are those that I

will use for the final experiment. This is for a number of reasons. Firstly,

metal weights are being used as the acting force and the smallest one that is

available to me is 50g which equates to 0.5N, and so this is the lower group

boundary. As for the upper boundary the value that I have chosen is 4.5N as the

next value above available to me, 5N, caused the trolley to accelerate at such

a rate that by the time it reached the end of the course, and the weights had

reached the ground, 20 marks had not been made on the ticker-timer paper and so

proper analysis could not be undertaken ( a fair test would not be maintained).Fair Test: A fair test must be ensured at all times, in any experiment,

to keep the results as accurate as possible so that appropriate conclusions can

be drawn. The main way that I hope to achieve this is by repeating each of my

results a further two times so that an average can be taken and any anomalous

results can be spotted before they are taken as genuine ones. As well as this I

must consider how accurate I want my results to be. As seen above I

think that giving my results to 3 decimal places would be appropriate as this

allows good continuity and does not suffer from premature approximation.

Another point that I will uphold is to use the same pieces of equipment for

every different interval. Should the experiments take more than one lesson then

I will mark each individual piece so I can recognise it at a later date.

Another point is the set-up must be the same for both experiments, if this does

not happen then I would not expect very accurate results at all. To make sure

that my results are accurate I will only change one factor at a time. In fact

there is only one factor that will be changed during the whole experiment. Factors To

Change Factors To Fix Factors to

Measure 1.The force being applied on the trolley (the total

amount of weights on the pulley system). Measured in Newtons. 1.The mass of the trolley, acting as the mass in

the formula F = ma. Measured in grams. 1. The acceleration of the trolley over the course.

Measured in metres /second². 2. The velocity of the trolley over the course. Measured

in metres / second. 3. The two different distances (see diagram). Measured in

centimetres. ? Safety Precautions: On the surface this is not a highly dangerous experiment,

however what must be shown is awareness of the environment that it is taking

place in. There will be many groups working within a very small area and this

means that conflicts can arise over space and working conditions. As well as

this the pull system will be dropping the weights into an aisle where other

groups of people will be walking. Therefore everyone will have to be vigilant

as to where they are walking. Other then this something soft will be placed

between the ground and where the weights will land. Plus the apparatus will be

kept securely on the bench so it is not knocked off with the clamp the clamp

being securely fixed. Hypothesis: Through my preliminary workings, and my initial scientific

research, I have begun to understand what I think my final results will show. Firstly my investigation is based around the formula of? F = ma. In the set-up that I am using

the only factor that is constant is the mass (the trolley). From this you can

tell something about the proportionality between the other two factors, the

force and the acceleration. That is that they are proportional, and this

is stated in Newton?s laws of motion. You can tell? proportionality on a graph because of two features: 1.The graph is a straight line ???????????????? 2. The line goes straight through the origin We

also know that when the graph has been drawn we will be able to take the

gradient of the line. This gradient should be equal to mass of the trolley:F = kak = F / a ??????? ??? ??????????? (k is an unknown constant, in this case the mass of ?the trolley or the gradient) If we were to keep the same force acting on the trolley, but

to change the mass each? time, this is

what we might expect.??????????????????????? m

= F / a ???????? ;?????????? m = k / a????????? ;?????????? m???

1 / aThis tells you that the mass of the trolley will be

proportional to inversion of the acceleration (1 / a). Yet again the graph of

this will be a straight line through the origin. From this, if you took the

gradient of the graph mass against 1 / acceleration then the

gradient will be the 1 / Force acting on the trolley. Here is how we know

this:??????????? m??? 1 / a ????????? ;

/ a??? = ????????? 1 / F ???????????????????????????????????????????? ?????????????? ??????????? ??m Overall I state that when Force is plotted against acceleration

then the graph will be directly proportional. Then if I take the gradient of

the graph then it should equal the mass of the trolley. On the other hand I

will only be able to speculate from my results what will happen when the force

is remaining constant as it will not be possible to have a range of masses to

do the experiment with, and the force is what I have stated I am changing each

time!Results: Mass /kg (trolley) Force /N (force applied) Distance 1 ????? /m Velocity1 /m/s Distance 2 ?/m Velocity2 /m/s Acceleration /m/s² Average Acceleration /m/s² 0.775 0.5 0.021 0.022 0.021 0.105 0.110 0.105 0.034 0.036 0.034 0.170 0.180 0.170 0.325 0.350 0.325 0.333 0.775 1.0 0.031 0.033 0.032 0.155 0.165 0.160 0.062 0.065 0.063 0.310 0.325 0.315 0.775 0.800 0.775 0.783 0.775 1.5 0.041 0.043 0.045 0.205 0.215 0.225 0.082 0.082 0.080 0.410 0.405 0.400 1.025 0.950 0.875 0.950 0.775 2.0 0.056 0.058 0.055 0.280 0.290 0.245 0.112 0.114 0.113 0.560 0.580 0.565 1.400 1.450 1.450 1.433 0.775 2.5 0.064 0.064 0.063 0.320 0.320 0.315 0.120 0.125 0.123 0.600 0.625 0.615 1.400 1.525 1.500 1.475 0.775 3.0 0.072 0.071 0.070 0.360 0.355 0.350 0.144 0.141 0.140 0.720 0.705 0.700 1.800 1.750 1.750 1.766 0.775 3.5 0.077 0.078 0.080 0.385 0.390 0.400 0.156 0.152 0.155 0.790 0.760 0.775 1.975 1.850 1.875 1.900 0.775 4.0 0.071 0.072 0.074 0.355 0.360 0.370 0.165 0.164 0.164 0.825 0.820 0.820 2.350 2.300 2.250 2.300 0.775 4.5 0.062 0.067 0.065 0.310 0.335 0.325 0.174 0.177 0.178 0.870 0.885 0.890 2.800 2.775 2.825 2.800 ??????????? ? Graph Analysis: Through my working I have been able to draw a graph of Force

against acceleration for which, as I have previously stated, the

gradient would be the mass. Although this is not exactly what I determined,

which I will come to later, I did find my initial hypothesis to be correct;

that is that as the force being applied to the mass (trolley) increases then

the subsequent acceleration of the mass (trolley) would also increase. However

there is more to it than that and so I will analyse all of my graphs more

closely. Graph A is a graph to show the average results for Force against

acceleration. The main point that I have focused on so far is that when

this graph is plotted, because of its direct relation to the formula ?F = ma, the graph shows the following

m = F / a. This means that when the gradient of the straight line is

taken (for this is a graph of proportionality) the gradient will equal the mass

in the formula (the trolley). I, indeed, did measure the gradient of the graph

(which should have been 0.775 ? 0.775g) and it did not equal the mass. In fact

with a measurement of 1.5, it is considerably far away, and if I were to take

it on face value then this would mean that the trolley weighed 1.5kg! Alarmed

at this I decided to draw another graph, which would show the results that I

had expected, and it was indeed quite different. There must be another force

acting that I have so far ignored, and yes there is. That force is friction,

which I have only briefly mentioned before in my preliminary work. Friction is

a force that opposes the movement of an object, and it acts in the opposite

direction to the way the object is moving. Between two surfaces it depends

on? a. the type of surface b .the

size of the reaction force. From these facts I can begin to understand why my

graph looks the way it does. Also, if looked at closely, the line of the

results does not go through the origin of the graph. This tells me that, just

like activation energy is needed to be overcome before a chemical reaction can

occur, a force is needed to provide an initial ?jump-start?. From this I can

say that the force needed to just start an object moving is equal to the

static friction value for the surfaces. This accounts for the ?error? in

reading at the start, but still there is an error in the overall gradient of

the graph. Therefore I can conclude that friction must be acting at all times

during the experiment (after all there is a straight line which means

consistency throughout the testing). A rule that I can draw from this is that the

force needed to keep an object moving steadily (with constant velocity) on a

surface is equal to the dynamic friction value for the surface. With this I

can account for the unexpected gradient, but without doing further experiments

all they are at the moment are theories (see further experiments).Conclusion: Overall my results were not as I would have expected them to

be, but I hope I have provided some insight into the reasons for this. From my

research I know that Force is proportional to acceleration, even

though my graphs do not show this but the reasons that I have given tell me why

they do not show it. This is because I did not anticipate the force of friction

acting on the experiment and if I had I would have taken measures to make sure

that they did interfere with my final readings, or if they did then I would be

able to account for them and tell what the experiment would have been like if

there was a frictionless environment. Just like activation energy is needed to

be overcome before a chemical reaction can occur, a force is needed to provide

an initial ?jump-start?. From this I can say that the force needed to just

start an object moving is equal to the static friction value for the surfaces. This

accounts for some of the ?error? in my results, but still there is an error in

the overall readings. Therefore I can conclude that friction must be acting at

all times during the experiment (after all there is a straight line which means

consistency throughout the testing). A rule that I can draw from this is that the

force needed to keep an object moving steadily (with constant velocity) on a

surface is equal to the dynamic friction value for the surface. With this I

can account for the unexpected gradient, but without doing further experiments

all they are at the moment are theories (see further experiments). My

experiments have left me with some conclusions that I can make: As

increasing forces are applied to a constant mass, the acceleration of the

mass also increases (F = ma). The

force needed to just start an object moving is equal to the static

friction value for the surfaces. The

force needed to keep an object moving steadily (with constant velocity) on

a surface is equal to the dynamic friction value for the surface. Accuracy of Results and how they relate to my original

hypothesis: On the surface the accuracy of my results was quite poor, on

the other hand I have accounted for the discrepancies that occurred. The only

reason that my results are not very accurate is that I did not account for the

friction in the system and if I had I?m sure that my results would have

supported the hypothesis that I put forward, and in a light they actually do.

There are ways that I could re do my experiment so that friction would not be a

problem and I have included some of the ideas later on. Evaluation: Overall I was quite pleased with what I have managed to take

from the experiment, not so much the results but the information, which I have

been able to take out of it. Although my results were the readings that I

expected to take, I was very happy indeed with the procedure and the way in

which I still managed to maintain fair conditions for it to take place. This

leads me on to the point that, although I did not take friction into account,

my results were still congruent and they still followed the pattern that I

expected and still followed the trends of the graphs that I included in my

hypothesis and preliminary work. This is shown by the fact that my best fit

line on graph a, despite having an inaccurate gradient, had the points plotted

very close to it. Also my readings did not show up any anomalous results, which

again, fills me with confidence if I ever repeat the experiment in the future

that my results would be accurate. Of course if I did indeed do the

experiment again I would have to take friction into account. The way in

which I would suggest to overcome this would be to use an air track (picture

included). Instead of using the ticker-timer (over a period of 20 dots) to

measure the acceleration, a series of light gates would be used in the same

way. This would completely rid the experiment of friction though due to it

being an air track there would still be some resistance from air molecules.

Though this method, if one does not already own an air track, would be an

expensive method. Therefore another method that could be used would be to make

the beginning of the course elevated from the finish. This could be done using

a beam that is propped up at the start end with item such as textbooks or a car

jack. The right would be that which compensates exactly for the friction in the

experiment. The main aim of my experiment was

to basically prove the theory of F = ma. The bottom line is that I could

not prove the proportionality of Force and acceleration, and my graph did not prove this as the

line, although straight did not pass through the origin. I hope that my

reasoning for this is correct and if it then I would brand my whole experiment

a success. On the other hand I would like to do the experiment again and

implicate some of the changes that I have suggested, and I know that the school

does own an air track so the results would be a lot more accurate.Further Experiments: The next experiment that I would put into action would be

either of the ideas that I have suggested in the last section so that my

overall results would be closer to those that I had expected. Also I would keep

the force acting on the trolley constant but change the mass of the trolley

each time to further investigate the formula of F = ma.? ? ?????????????????????????????????

ДОБАВИТЬ КОММЕНТАРИЙ  [можно без регистрации]
перед публикацией все комментарии рассматриваются модератором сайта - спам опубликован не будет

Ваше имя:

Комментарий

Хотите опубликовать свою статью или создать цикл из статей и лекций?
Это очень просто – нужна только регистрация на сайте.

## Видео

Calculating Force Mass Acceleration Part 3 of 3  [ВИДЕО]A Level Physics Force Mass Acceleration and Newton 39 s Second Law  [ВИДЕО]
 ● Newton 39 s Second Law Force Mass and Acceleration  [ВИДЕО] ● GCSE Physics Revision Force and Acceleration  [ВИДЕО] ● Tension In Rope Between Two amp Three Blocks Accelerating System Physics  [ВИДЕО] ● Calculating Acceleration with Friction Net Force Horizontal  [ВИДЕО] ● 040 Force mass x acceleration example 1  [ВИДЕО] ● Net Force mass and acceleration examples  [ВИДЕО] ● Force mass acceleration example spring compression  [ВИДЕО] ● A Level Physics Force Mass Acceleration and Newton 39 s Second Law  [ВИДЕО] ● YouTube Newton s Laws Of Motion 2 Force Mass And Acceleration  [ВИДЕО] ● force mass and acceleration formula  [ВИДЕО]