To find out what happens to the efficiency of a motor as I change the mass it lifts

When devices transfer energy, only part of it is USEFULLY TRANSFERRED to where it is wanted and in the form that it is wanted. The rest is transferred in some non-useful way and therefore it is ‘wasted’. The ‘wasted’ energy and the ‘useful’ energy are both eventually transferred to the surroundings. The greater the proportion of energy supplied to a device, THAT IS USEFULLY TRANSFERRED, the more efficient we say the device is.A motor is a device that transfers electrical energy into rotational kinetic energy, which can be used to lift a load. We are going to try and find out how the efficiency of a motor differs as we change the mass that it is required to lift. To do this we will let a small electric motor lift a small load 0.5m off the ground and work out it’s efficiency, increasing the weight of the load it has to lift by 0.1N each time we run it. Below is a diagram showing how the circuit for this experiment will be set up:As you can see, the motor has to be connected to the ammeter, voltmeter and the power supply. The ammeter is placed in series and the voltmeter is placed in parallel. The motor should be clamped tightly onto a stand over one metre off the ground. A piece of string capable of reaching to the floor should be attached to the spindle of the motor, whilst the other end should be attached to the mass hook.When the experiment is run, a stopwatch should begin timing as the power supply is switched on. Whilst the load is being lifted the amps should be read from the ammeter, and the volts from the voltmeter, at the same time. Once the load has been lifted 0.5 metres the timing should be stopped and the data recorded: weight (N), volts (V), amps (A), time (s). The basic raw data from the experiment has now been obtained. This process is repeated twice for each different weight that is lifted, amounting to a total of three runs per change in weight.The purpose of the volts, amps and time readings is to calculate the electrical energy supplied to the motor as it runs, or the total energy input, required later for calculating the efficiency. To calculate the total electrical energy supplied to any device we need to use to use the following formula:ELECTRICAL ENERGY (J) = VOLTS (V) ? AMPS (A) ? TIME (s)The volts, amps and time values were multiplied together to calculate the total electrical energy supplied to the motor in our experiment, as you can see in Table #1 in the Tables of Data section in the column headed Electrical Energy.PredictionThere has to be a weight that a motor is incapable of lifting. So although the motor can do more work by drawing more current, it must become less efficient.I predict that the line of best fit showing the trend on my results graph will look like:In other words, as the mass of its load increases, the efficiency of the motor will decrease by a uniform rate.I predict this because the heavier the weight, the more rotational kinetic energy will be required to turn the motor’s ‘spindle’ to lift the weight. An increase in the amount of rotational kinetic energy being transferred must also result in:* An increase in friction affecting the motor’s moving parts, causing energy to be wasted in the forms of heat and sound.* An increase in the current drawn by the motor. This causes an increase in the resistance of the wires connecting the motor to the power supply, which in turn causes electrical energy to be wasted in the form of heat, lowering the amount of useful energy output, and therefore lowering the motor’s efficiency.Calculating EfficiencyIn order to find the motor’s efficiency each time we shall have to calculate several values. Firstly we need the raw data obtained from the experiment:* Weight Lifted (N): the weight the motor is lifting in Newtons.* Height (m): the height in meters that the load is lifted – always 0.5m.* Volts (V): the voltage taken from the voltmeter in the circuit.* Amps (A): the number of amps read from the ammeter in the circuit.* Time (t): the time in seconds that it takes the motor to lift it’s load 0.5m (50cm).We then need:* Average Electrical Energy (J): the average electrical energy supplied to the motor in Joules.* G.P.E. (= Movement energy) transferred (J): this is the useful energy output we get from the motor in Joules.Which allows us to calculate:* Efficiency (as a value between 0 and 1): the efficiency of the motor based on all the other values.On to a graph I can then plot:* Weight/Newtons LiftedAgainst* EfficiencyRESULTSTables of DataBelow is the data that I used to calculate the motor’s efficiency, set out in three tables of values, though the calculations used to convert the values are explained later. All of the numbers are given to two decimal places.Note: The calculations for these values will be explained later.Table #1: This table displays the three measurements of electrical energy taken each time the motor was run, and the weights it was required to lift.* = counted as an anomalous result – not included when averaging dataElectrical Energy (J)Weight (N)0.330.310.270.100.75*0.300.360.200.600.590.530.300.490.790.590.400.710.660.910.501.20*0.790.830.601.001.311.240.700.971.32*0.870.801.781.861.44*0.902.532.703.47*1.00Table #2: This table displays the average electrical energy supplied to the motor, the change in G.P.E. of the motor (the useful output) and the efficiency of the motor worked out from the other values.Average Electrical Energy (J)G.P.E.=Movement Energy (J)Efficiency0.300.050.170.480.100.210.570.150.260.620.200.320.760.250.330.810.300.371.180.350.300.920.400.431.820.450.252.620.500.19Table #3: This table displays the values to be plotted onto my results graph: Newtons lifted (weight of load) against efficiency.Newtons Lifted (N)Efficiency0.100.170.200.210.300.260.400.320.500.330.600.370.700.300.800.430.900.251.000.19ANALYSING AND CONCLUDINGBelow is a diagram of a running motor. An explanation of what is happening is given below.1. Electrical energy is supplied to the motor.a. Some of this electrical energy is transferred into the desired movement (rotational kinetic) energy in the motor – this is useful energy.b. Some of the electrical energy is transferred into non-useful thermal (heat) and sound energy – this is waste energy.2. The waste energy is lost to the surroundings.3. In this case the movement energy is now transferred into gravitational potential energy when the load is lifted.Energy has to be transferred from one form to another, e.g. a hairdryer turns electrical energy from a mains supply into movement energy (the fan), heat energy (to heat the air as it passes through) and sound energy (waste energy). You cannot create it or destroy it. Energy efficiency is how much of the energy you put into an appliance or machine is transferred into the useful energy that you are trying to get out. All machines in the real world have an efficiency that is less than 1 (or 100%). In the case of the motor above, part of the electrical energy put in is transferred into the useful movement energy, however, the machine also transfers it’s energy into two other waste forms: it creates a little heat and a little sound, caused by the force of friction on it’s moving parts, as in all machines. The greater the proportion of energy supplied to a device, that is usefully transferred, the more efficient the device.To calculate the efficiency of any device we need to use to the following formula:EFFICIENCY = USEFUL ENERGY TRANSFERRED BY DEVICETOTAL ENERGY SUPPLIED TO DEVICE… So in the case of an appliance that coverts 200 joules of electrical energy per second into 150 joules/sec of waste heat energy, 20 joules/sec of useful light energy and 30 joules/sec useful sound energy… EFFICIENCY = 50 ? 200 [? 100] = 25%We are trying to find out how the efficiency of a motor changes as we increase the mass it has to lift, so we need to remember that the efficiency of a motor is determined by how much of the electrical energy put in is transferred into useful energy output. The useful energy output is gravitational potential energy (GPE). Gravitational potential energy is the ENERGY STORED in an object because of the HEIGHT that the WEIGHT (due to the force of gravity on the object) of the object has been lifted against the force of gravity. If an object can fall, it’s got gravitational potential energy. The motor we used for our experiment was set up to lift a load 0.5 metres off the ground, in other words, the rotational kinetic energy of the motor’s spindle is transferred into movement energy (the same type really) as the load lifts, and by the time the weight has been lifted to it’s full height, all the movement energy will have been converted into gravitational potential energy, as the load can now fall. Because calculating efficiency requires that we know the useful energy output, we shall need a way of calculating the change in GPE of the load once it has been lifted (how far the weight has moved). To calculate the change in gravitational potential energy we use the following formula:CHANGE IN GRAVITATIONAL = WEIGHT (N) ? CHANGE IN VERTICAL HEIGHT (m)POTENTIAL ENERGY (J)… So in the case of an object weighing 800N undergoing a change in height from 1000m to 3000m above ground, the GPE of the object can be worked out using the formula:CHANGE IN GRAV. POT. ENERGY = WEIGHT ? CHANGE IN VERTICAL HEIGHT= 800N ? (3000m – 1000m)= 800N ? 2000m= 1600000JIn our experiment we calculated the GPE (useful energy output) like this:GPE = Weight of motor’s load in Newtons ? The height it was liftedThe height that the load was lifted remained the same throughout the whole experiment. The weight lifted was incremented by 0.1N to find how this affected the efficiency. The values for gravitational potential energy are displayed in Table #2 in the Tables of Data section, along with the averaged values for the total electrical energy supplied to the motor for each weight change.With these two values: Electrical Energy and Change in GPE, we have the necessary data needed to calculate the efficiency of the motor. To calculate efficiency, we are required to use the formula:EFFICIENCY = USEFUL ENERGY TRANSFERRED BY DEVICETOTAL ENERGY SUPPLIED TO DEVICEWe can now substitute the two values needed to find the efficiency of our motor with our own data from the experiment, where GPE is the useful energy transferred and Average Electrical Energy is the total energy supplied:EFFICIENCY = GPE (J) ? Average Electrical Energy (J)… Giving us the values shown in the column headed Efficiency in Table #2.Graph and Line of Best FitI have drawn a graph showing the results that I have collected. The correlation of the points made it possible to draw a line of best fit onto my graph. By looking at the graph it is possible to determine a clear trend, to find the optimum weight for the motor’s efficiency, and to spot any anomalous results.What I have found outBy looking at the graph you can see that the line of best fit shows a clear relationship between the weight of the load and the efficiency of the motor: at first, an increase in the weight of the load causes an increase in the motor’s efficiency. This is only true up until the weight reaches 0.6N, at which point the efficiency peaks, and the efficiency decreases as the weight continues to increase.I now know that the efficiency of a device does not remain constant, it is affected in some way by the ‘work’ that it has to do.The gravitational potential energy is affected only by the change in mass (or the affects of a change in mass), as both the electrical energy supplied to the motor and the height the load had to be lifted were unaltered by us.In summary, I have found the efficiency of the motor increases whilst lifting relatively light weights, until the highest efficiency value is reached, at which point it begins to decrease.How close is this to my prediction?I predicted that the efficiency of the motor would decrease by a uniform rate as the weight of its load was increased. I predicted this because I already knew that there must be a weight a motor is incapable of lifting, so its efficiency must be affected by the weight of it’s load. I also drew a sample of what I thought my line of best fit would look like. This turned out to be partially correct: the real line of best fit clearly shows the efficiency decreasing, but only after an initial increase before reaching 0.6N.My graph backs up these comments. As you can see, the weights tested that were higher than 0.6N produce a clear downwards trend, showing that the efficiency of the motor decreases as the weight of its load is increased.Explaining what I have found outThe first trend – why does the efficiency increase?My results graph tells me that an increase in the weight of our motor’s load results in an increase in efficiency. In other words, the lighter the load, the lower the efficiency (up until 0.6N) – why is this? I believe it is due to the fact that a light load does not produce enough tension in the string attaching it to the motor to prevent the string from slipping on the spindle. This would have made it appear on my graph that the motor itself was inefficient, when in fact it was due to the string gripping the spindle improperly. The string slipping was caused by a lack of friction between the string and spindle, resulting in less rotational kinetic energy in the spindle being converted into gravitational potential energy – the efficiency of the motor appeared to be lower for lower weights, though in actual fact the energy output was altered by the string slipping, rather than by the efficiency of the motor. The reason for the initial rise in efficiency was that the greater the weight between string and spindle, the more friction, therefore the less slipping occurred, causing the gravitational potential energy (useful energy output) to be higher, regardless of the motor’s efficiency.The second trend – why does the efficiency decrease?My line of best fit shows that increasing the weight of the motor’s load beyond 0.6N causes a decrease in efficiency. There are several reasons explaining why an increase in weight causes efficiency to begin decreasing.FrictionFriction is a force that is created whenever two surfaces move or try to move against/across each other, i.e. the moving parts inside an electric motor. Friction always opposes the motion or attempted motion of one surface across another surface.1. Friction between the string and motor spindle: Our motor lifts its load by transferring the electrical energy supplied to it into rotational kinetic energy in the spindle. This winds the string causing the load to move, until all the movement energy has been transferred into gravitational potential energy. If we increase the weight of the load, there is increased tension in the string. This means that there is increased friction where the string is attached to the spindle, in other words, as the spindle turns, there will be more resistance between it and the string. The friction between these two surfaces results in heat and a little sound being produced. Because we know that energy can be changed from one form to another but cannot be created or destroyed, we know that the heat and sound energy produced as a result of friction between the string and spindle must therefore have been transferred from the electrical energy inputted. Both heat and sound energy are non-useful to us in this case, therefore the more electrical energy transferred into these waste forms, the greater the apparent reduction in the motor’s efficiency. It is important to note that it is friction between the spindle and the string, not the efficiency of the motor that is responsible for a decrease in useful energy output in this instance. Though the graph makes it appear that the motor’s efficiency is decreased, the actual rotational kinetic energy transferred is not affected by this friction (unlike point 2 below), since it is during the intermediate stage, when the energy in the spindle is transferred to the load, that the GPE (the useful energy output that we measured) is decreased.2. Friction affecting the moving parts inside our motor: The heavier the weight, the more movement energy the motor’s ‘spindle’ will need to have in order to lift the weight. The motor is able to draw more current allowing it to transfer a greater amount of rotational kinetic energy. An increase in the amount of rotational kinetic energy being transferred to the spindle (it has more ‘spin’) must also result in an increase in friction affecting the motor’s moving parts. This friction creates heat and a little sound, which can only have come from the electrical energy supplied to the motor, meaning that energy is ‘wasted’, lowering the motor’s efficiency. Essentially, increasing the weight of the load causes an increase in friction therefore decreasing the efficiency. This explains the decreasing trend we see on my results graph.ResistanceComponents resist the flow of current through them. They have resistance. Electric current is the movement of electrons through a conductor. Metals (e.g. wires) conduct electricity because the atoms do not hold on to their electrons well… FREE ELECTRONS are present. The number of mobile electrons that are in a length of wire will produce a certain amount of resistance. This is because when current is passed through the metal the electrons of the particles are given energy allowing them to move. As they travel through the wire, they come in to contact with impurities and other particles, which they ‘bump into’. This collision releases some of the electrical energy as heat energy, which is lost to the surroundings. The greater the current, the more electrons flow. The more electrons flow, the more collisions occur, causing a greater amount of the electrical energy to be wasted as heat.In the case of our motor:* When we increase the weight of its load, we increase the amount of ‘work’ the motor must do, or how much rotational kinetic energy must be transferred.* The motor is able to draw more current to allow it to do more work.* An increase in the current flowing through the motor’s wires…* … Causes an increase in waste heat energy…* … Which can only have come from the electrical energy supplied to the motor in the first place, therefore lowering the efficiency.Vibrating motorWhen motors ‘work hard’ they have a tendency to vibrate or shake slightly. I believe this is due to conflicting forces inside the motor. When we increased the weight our motor had to lift, we caused these forces to become stronger, therefore the motor would have vibrated or shook more violently. This kinetic energy has to come from somewhere – we know it must have been transferred from some other form, as energy cannot just be created… It must have come from the electrical energy inputted. I now know that the motor vibrating contributed lowering the efficiency of our motor because an increase in the weight of it’s load causes more energy to be transferred resulting in greater conflicting forces, making the motor vibrate, or lose energy as non-useful (waste) kinetic, decreasing the efficiency of the motor.0.6N – why does it peak here?My line of best fit peaks at 0.6N, showing this to be the ‘optimum weight’ for maximum efficiency for this particular motor. This must be the point at which the efficiency stops increasing, and begins to decrease. In other words, the amount of non-useful ‘waste’ energy stops decreasing, and begins to increase instead. So why does the first trend cease? I believe that a load weighing 0.6N must have produced enough tension in the string, and therefore enough friction to grip the spindle of the motor effectively, allowing the motor to appear at it’s most efficient. Weights exceeding 0.6N however, caused increased friction affecting all of the motor’s moving parts, producing two waste forms of energy: heat and a little sound, enough to start lowering the motor’s efficiency. A large enough increase in resistance and vibration of the motor were also factors that contributed to a decrease in efficiency.EvaluatingAnomalous ResultsIn my results table containing raw data from our experiment (table #1) I have marked any results that I considered to be anomalous with an asterix (*). These anomalous results were not averaged along with the others.On my graph I have circled what I consider to be a single anomalous result.I consider these results to be reliable because they do not ‘fit in’ with the rest. Values marked with an * are not the values you would expect to come alongside the other two results from that particular weight tested. For example, take the three original values for electrical energy supplied to the motor when lifting 0.20 Newtons:0.75*0.300.36You can clearly see that 0.75 does not match with the other two results, nor does it fit in with what you might expect it to be, considering the results for other weights – it is not part of the trend. It has therefore been marked as anomalous and not including when averaging.Are my results reliable?Indications of a reliable experiment are:1. Results that, when plotted on a graph, form a very clear trend. That is, a line of best fit can easily be determined, and many points fall on it.2. Results that, when plotted on a graph, do not present [many] anomalous results – results that do not obey the general trend, which must be discarded/ignored when drawing a line of best fit.3. Results which, when an experiment is repeated, are close together for repeats. E.g. 0.32, 0.33, and 0.34 would be considered reliable in our experiment, as these could be averaged confidently.Bearing these points in mind, it is possible to judge the reliability of my results.When I plotted my final results onto a graph, I had to draw a line of best fit – a line that best took account of the trend of all the valid results plotted. My line of best fit was difficult, but by no means impossible to draw. The points did demonstrate a trend, which could easily be determined, but it was not possible to know exactly where to draw the line. The points are far too loosely arranged, and there are to few of them, to provide a strong enough trend to indicate an unarguable line of best fit. An unquestionably reliable set of results could look like this:However, my points merely suggested the shape of a line of best fit. Also, the sketch above shows many points falling directly on the line of best fit. My line of best fit has no points exactly on it, which shows how difficult it was to draw – I was required to guess how the line would pass through the gap between two points, etc.The number of anomalous results in the data and on the graph can also show how reliable the results are. My original data for electrical energy in Table #1 contains quite a few anomalous outcomes, indicating that the results of my experiment are dubious. Things that are reliable tend to happen the same way time and time again, so a completely reliable experiment would produce very similar results for any repeats. Although my results contain several severely anomalous results, there not a great cause for concern, as the repeats that are not anomalous definitely have relationship to each other. They are close enough to be averaged reliably when the anomalous results have been omitted.As for anomalous results on the graph, there is only one clearly anomalous outcome in my opinion, which is the efficiency for the load weighing 0.8N. If all the results were this scattered, the results would be completely unreliable, but it is easy to recognise as anomalous, and the rest of the points still provide a strong enough pattern as to imply a trend and line of best fit.In summary, my results are quite unreliable, but are still useful and form a good trend. The graph agreed in part with my prediction, and can be explained using science. This shows that the results were fairly accurate and reliable, and although results were present that were obviously incorrect, they were easy to spot and eliminate. However, the points on my graph were quite scattered and ‘loose’ showing that my results could have been more accurate.Do I have enough evidence to support a conclusion?In my conclusion I tried to explain the results that I obtained. I believe that I do have enough evidence to support my conclusion, though I would have preferred much more data, and if possible collected in a more accurate way, as this would have made my conclusion far easier to back up – the more evidence I have the more sure I can be about the statements I have made, and the larger the amount of data I have to draw upon and use to suggest a trend. This would have made the experiment more worthwhile, and the results far more reliable.I know I have enough results because it was possible to draw a line of best fit, and guess at how the trend would have continued. However, a good example of how I would have liked more evidence is at the very top of my line of best fit, where I have suggested the optimum weight for maximum efficiency of the motor to be. With our current set of results, we can not really be sure where the layout of further points would be around that area, and exactly how the line of best fit should be shaped. For instance, the optimum temperature could have been at a slightly different point, because I was unable to tell exactly when the line of best fit should begin to slope downwards when I drew it.Using science it is also possible for me to suggest what further results would be without doing further experiments. For example, I can predict how the line of best fit would be shaped for lighter weights.How well did I carry out the investigation? -Problems we had that affected the resultsDuring the course of the experiment, we encountered several problems that may well have affected my results.1. The ammeter and voltmeter readings: The ammeter and voltmeter readings were both taken at a given time whilst the motor’s load was being lifted. The displays were fluctuating, showing the readings to be inaccurate. Also, the two people taking the readings found it difficult to judge an accurate value, because of the changeability of the readings.2. Timing for lighter loads: Whilst testing lighter weights, the load was lifted too fast for an accurate measurement of time to be taken, using a simple stopwatch. Therefore the timing would have been inaccurate for some of the lighter weights tested.How fair was our test?I think that our experiment was very close to being as fair as possible, considering the time and equipment possible, because we considered all the things that might have made the results unreliable, even if we could not control them. We made sure to keep all of the runs the same by conducting them under exactly the same conditions, so we can be relatively confident about the accuracy of our results. However, the results are definitely unreliable, though this is mostly due factors beyond our control which shall be looked at later. Below are the ways in which we attempted to keep our experiment a fair test, thus increasing the accuracy and usefulness of our results.What we did to make it a fair:Using the same people to take time the experiment and say ‘Now’: We made sure to only the same two people for the following two jobs:1. Starting and stopping the stopwatch at the beginning and end of each run so that the time taken for the load to be lifted could be measured.2. Saying ‘Now’ to indicate to the two other people reading the ammeter and voltmeter when the values should be taken.We made sure not to swap these people around because maintaining the accuracy of our results relied upon both the reaction time and hand-to-eye coordination of these two people. Changing them around would have meant using people with different reaction times, so the test would have been unfair.Conducting the experiment on just one occasion and not repeating on a different day: The whole experiment was conducted on a single occasion. We did not allow the experiment to take place over several days, because the equipment we would have used would have been different, causing unfairness in the results. For example, we almost certainly have used a different motor, which, although it might look similar, would have been slightly different than the first, as they are not precision made. The efficiency for this motor would have different and therefore the data as a whole would have been unreliable.When we first looked at our results graphs, a clearly anomalous result was apparent (this point has been ringed on my graph). Although it was tempting to go back to the experiment, set everything up as close as possible to before and test that weight again, we did not do this. If we had, different equipment would have made our results inaccurate. The resistance of the wires used and of the motor would have been different, as length and amount of rust or damage can alter this. Also, we had no guarantee of using the same motor again, which is vital when testing efficiency.Calculating averages and omitting anomalous results: When carrying out the experiment, we made sure to test each weight of load three times, recording the electrical energy input for each repeat. The experiment threw up several anomalous results that can be seen in the raw data. When we calculated an average electrical energy value for every three repeats we omitted any anomalous results. These were results that were unexpected, and gave clear signs that something went wrong during that run, e.g. a ‘hiccup’ in the motor’s efficiency, or a misread ammeter value. By omitting these results, I was able to give a fairer set of results, consequently increasing the reliability and ease with which my results could be analysed.Keeping the height from which the load started from the same each time: To keep the amount of electrical energy needing it be transferred into gravitational potential energy the same for every lift, we made sure that the motor’s load was place at exactly ground level, with the string taut every time the experiment was run. If the load was above ground level, it would have taken less energy to be transferred to lift it 0.5m, giving an unfair picture of the efficiency. We made sure that the sting was taut each time, because if it had been slack, the motor spindle would have had to complete more rotations in order to wind the string and lift the weight.What may have caused anomalous results?When we conducted the experiment everything went relatively well and according to plan. We did everything we could to keep it a fair test and tried to control all the factors that might have made the results inaccurate. However, there were many things that could have caused inaccuracy that were beyond our control, some of which were:1. Reaction time/human error: The accuracy of the results relied heavily on the consistency of the timings, carried out by someone stopping and starting a stopwatch when the power was turned on and when the load had been lifted 0.5m. However, it is impossible to time an experiment completely accurately using such basic equipment, and without letting human error affect the results. Slight (but nonetheless important) imprecisions will have occurred when timing how long it took the motor to lift its load, causing the results to be inaccurate and unreliable.2. The way in which the string wrapped around the spindle: The string could be seen winding round the motor’s spindle in different ways. This would have affected the amount of friction between string and spindle, which could have both lowered the efficiency or increased it.3. Motor efficiency being altered of it’s own accord: The motor itself may have caused anomalous results. Sometimes it seemed to run very slow, at other times very fast, regardless of what weight it was lifting. This could have been due to the motor drawing too much current, or internal parts functioning irregularly or not as they should.4. Ammeter and voltmeter readings: The ammeter and voltmeter readings were fluctuating, so it is doubtful if they are accurate. This could also mean that the motor alters the amount of current it draws as it lifts its load. This means that a reading taken at one given point will not give an exactly fair representation of the electrical energy being inputted. In an attempt to combat this problem, we read the ammeter and voltmeter when the load had been lifted exactly half way. This however, could have caused the results to be inaccurate for a different reason, as it relied upon someone saying ‘Now’ when the load had been lifted 0.25m off the ground. This means the results will have been affected by human error and reaction time – how soon after the weight was lifted 0.25m was the indication given to read the ammeter and voltmeter? How quickly could the values be read after shouting ‘Now’? The imprecisions involved could have caused the results to be inaccurate or anomalous.How could I improve the investigation?If we conducted the experiment again, it could be improved in the following ways:1. Obtain more results: By this I do not just mean testing weights higher than 1.0N and lower than 0.1N. We could obtain efficiency values for weights incrementing by 0.05. This would allow me to determine a better trend, and draw a more accurate line of best fit. By obtaining more results, I could be more confident about my conclusions and have more evidence to back up my statements.2. Use a more accurate timing device: Because our results would have been affected by the inaccurate methods of timing, we could attempt setting up a circuit that uses pressure pad to determine exactly when the load is lifted 0.5m and how long it takes – the timer would be activated when the load started to be lifted and stop when it had been lifted to the height required. This would prevent human error affecting our results and causing them to be less reliable.3. Use an electronic ‘joule meter’ in our circuit: Our experiment used values taken from both an ammeter and voltmeter and a stopwatch to calculate electrical energy supplied to the motor using the V ? A ? T formula. However, because of the fluctuation and difficulty in reading the ammeter and voltmeter, as well as the imprecise method of timing the experiment, the results were made considerably less reliable as a result. By simply using a joule meter in our circuit to tell us the electrical energy inputted we could eliminate all need for Volts, Amps, and Time readings. By replacing these inaccurate pieces of equipment we could increase the exactness and reliability of our results, making the experiment more worthwhile.What Further Work Could I Do to Collect Further Evidence?If I had more time there are other things that I would do to collect extra evidence, that is relevant to this investigation:Testing weights below 0.1N and above 1.0NWe could test loads that were lighter than 0.1N to see how the trend starts – is it straight or does it tail off slowly? This might prove difficult however, as very light weights might be too difficult to time accurately using our method, or product ammeter and voltmeter readings that are too small to read or to work with. If we tested weights above 1.0N I believe we eventually find a weight which the motor would be unable to lift. At this point, there would be no useful energy being transferred, all electrical energy inputted would be transferred into non-useful waste forms. It would have been interesting to find out what this weight is. My graph displays a line of best fit which makes it difficult to predict how it will continue. It is very hard to tell from my data whether the line should slope downwards in a straight line, or decrease slowly at first, then more quickly, etc. Therefore collecting more data would provide me with more informative results.Carrying out further tests around 0.6N – the apparent optimum weightIt would be interesting to try and find out exactly where my line of best fit should peak, and what the precise optimum weight for the efficiency of this motor is. By collecting further, accurate results I could determine what weight this motor will lift most efficiently, e.g. 0.62N.By plotting more points around the area where my line of best fit peaks on my graph, I would be able to tell exactly how my line of best fit should be shaped around it’s peak.Lifting the loads higher and finding out at what point is a motor most efficientWe could try and find out how much a motor’s efficiency differs depending on how much of the way through a lift it is. For example, is a motor more efficient when it has just begun lifting it’s load, or when it is almost at its full height. We could do this by letting a motor lift a load slowly, whilst measuring the electrical energy input just after it starts to lift, and whilst it has been running for longer.We could also try and determine whether a motor is on the whole more efficient when lifting to a small height or to a large height, or in other words, how does the length of time it has to lift a load for affect a motor’s efficiency?