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I believe that it is appropriate to begin this coursework with a quote Essay

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Paper type: Essay, Subject: Free

Every body continues in its state of rest or of uniform speed in a straight line unless it is compelled to change that state by forces acting on it.Newton’s First Law of Motion (Law of Inertia)This quote, Newton’s first law of motion, is basically what I am trying to prove. The idea that an object will continue to move with the same speed unless some other force is acting on it. In this case, friction.Account of the Experiment and Data Gained.The data that I am going to analyse has been obtained from an experiment performed in class, involving a buggy being rolled down a slope at an incline of 7.7o and passing through a light gate which I shall move down the ramp gradually. The diagram below (Fig 1)illustrating the experiment.Fig 1The experiment was performed for eight different range’s 20cm – 160cm, increasing in steps of 20cm with 3 attempts at each distance.Plan of calculationsThe aim of my coursework is to attempt to discover the coefficient of friction and locate any patterns throughout the data.To look for patterns I will plot a distance time graph and a force time graph, working out the gradient to give me acceleration down the ramp. I will also work out the gravitational component with expected values.To attempt to find the coefficient of friction, I will look for any deviance from the expected values of acceleration, and once the results are compared, discover the differing value of friction.I will use the equations F = Ma and a = ?V/?T.Expected ValuesFig 2To find velocity A I will break up the downward force C. I know the angle of the ramp isSin = o/h. =0.327/2.44 = 0.134. Sin-1 = 7.701oThe other angle (B) in the triangle = 180-7.701-90 = 82.3oSo breaking up the force gives us: -Fig 3Original force (Black) of 9.8 x 0.6678 = 6.544(The error in the mass is extremely small+/-0.0001, I have decided that it is negligible in the following calculations)Force A (Blue arrow) of 6.544cos82.3 = 0.877And Force B (red) arrow that is balanced by the upward force of the ramp(Green upward arrow).Acceleration (at bottom of ramp) = F/M = 0.877/0.6678 = 1.3 ms-2The potential energy of the trolley at the top of the hill should be 0.877N, and get to the bottom with an acceleration of 1.3 ms-2. The acceleration values from the graph need to be below 1.3 ms-2 to be acceptable data.Data analysisData that was gained from the experiment: -Fig 4DistanceTimeVelocityRun 1Run 2Run 3AverageRun 1Run 2Run 3Average200.410.40.310.3730.720.730.710.720400.590.590.630.6031.021.021.021.020600.780.780.910.8231.231.221.231.227801.091.1511.0801.391.391.391.3901001.091.151.121.1201.541.541.541.5401201.381.411.281.3571.641.641.631.6371401.431.51.561.4971.731.731.721.7271601.531.651.841.6731.841.841.831.837To calculate the average acceleration, I used a = ?V/?TFig 5Average Acceleration01.9291.6911.4901.2871.3751.2061.1541.098As you can see. The average acceleration of the cart is way over the 1.3 ms-2 that I estimated above. This could possible be explained by the cart having been pushed and so has an initial velocity ; 0.The most useful graph to plot will be a Velocity/time graph, (Graph 1), using this to calculate acceleration, from the acceleration values I will be able to work out the resistance value: -(Resultant) F = Mg-Fr(Since F= Ma)Ma = Mg-FrBy rearranging this formula to get the Frictional Force Fr.A Fr = Mg-MaSo by working out lots of acceleration values I will gain resistance values. The values have been taken from graph 9.(Fig 6)1. Into formula AFr = (0.6678 x 9.8 x cos82.3) – 1.15 x 0.6678Fr = 0.876 – 0.768Fr = 0.1082. Into formula AFr = (0.6678 x 9.8 x cos82.3) – 0.2 x 0.6678Fr = 0.876 – 0.134Fr = 0.7423. Into formula AFr = (0.6678 x 9.8 x cos82.3) -0.08 x 0.6678Fr = 0.876 – 0.053Fr = 0.823The next graph will be a Friction / Distance graph (Graph 2). From it I hope to establish if the frictional force is constant or changes.The distances I will use will be – 20 cm, 60 cm, 100 cm and 140 cm. I shall use the data gained in fig 5 for this graphFig 7a = ?V/?TFor 20cma = 0.72 – 0/0.373 – 0a = 1.929 ms-2For 60 cma = 1.227/0.823a = 1.490 ms-2For 100 cma = 1.540/1.120a = 1.375 ms-2For 140 cma = 1.727/1.497a = 1.154 ms-2From this data I will calculate the expected friction using the formula: -Ma = Mg-FrFr = Mg – MaAv AccelerationMassGravityFriction00001.9290.66789.85.2571.6910.66789.85.4151.4900.66789.85.5491.2870.66789.85.6851.3750.66789.85.6261.2060.66789.85.7391.1540.66789.85.7741.0980.66789.85.811From this graph (Graph 2) I have determined that as distance increases so does the resistance. The resistance increases in regular amounts so I will be able to work out that amount to give me the coefficient (coeff) of frictionNow I have values for resistance I will attempt to find a relationship that will allow me to find the coefficient of friction. The frictional force depends on 2 things, the coefficient of Friction with the surface that the cart is in contact with and the upward force exerted by the ramp, which I will call Fup. For this formula to work I must treat the cart as a point mass.The formula that I have found is Fr = coeff Fr x FupThe friction data was taken from fig 6.Inserting the data for gives me0.108 = coeff Fr x (6.544 x cos84.2)Rearranging gives me an answer of: -0.108/0.865 = 0.125By subtracting this value from the perfect results I should find the actual force of acceleration.0.877-0.125 = 0.752Now by using a = F/M I should find the acceleration that I found in part 1).a = 0.752/0.6678a = 1.126 ms-2This value is extremely close to the value of 1.15 ms-2 from the graph and I put down the difference to inaccurate gradient measurement. The alternative to this would be that the cart actually had 2 coefficients of friction. One to start the cart, the other is whilst the cart is moving.Comparison between the gravitational potential energy and the energy transformed into kinetic energy.K = 1/2mv2This is the formula that I will use to calculate the kinetic energy transferred to the cart. With the following equation I will work out gravitational potential energy.?GPE = Mg?hTo start I will calculate the kinetic energy.Starting with the following data I will perform a sample calculation.Av VelocityMass0.7200.6678K = 1/2mv2= 1/2 x 0.6678 x (0.72)2= 0.173 JThe rest of the data follows suit: -kinetic energy (J)Av VelocityMass0000.1730.7200.66780.3471.0200.66780.5021.2270.66780.6451.3900.66780.7921.5400.66780.8941.6370.66780.9951.7270.66781.1261.8370.6678Next I will work out gravitational potential energy?GPE = Mg?hTo work out the height at each of the distance values I will use trigonometry.Sin????a?hSin7.7o = a/0.2a = 0.2 x sin7.7a = 0.03I shall use the distance values from the following points.Here is a sample calculation. For a ramp length of 0.2 and height 0.03GPE = 0.6544 x 9.8 x 0.3= 1.309Distance (M)Mass (Kg)GravityGPE00000.20.65449.81.3090.40.65449.82.6180.60.65449.83.9270.80.65449.85.23610.65449.86.5441.20.65449.87.8531.40.65449.89.1621.80.65449.810.471Comparing Kinetic energy and gravitational potential, shows that a tiny proportion of energy is actually converted into kinetic energy as the table below and graph 3 shows: -GPEKinetic Energy001.2830.1732.5650.3473.8480.5025.1300.6456.4130.7927.6960.8948.9780.99510.2611.126ConclusionIn conclusion the cart is extremely inefficient and looses lots of potential energy through friction. There were many problems with the data, there was flex in the ramp, the measurements for distance were very rough and we did not use a set square. The light gate was only lined up by eye and there were parallax errors in the time calculations.From the data I have plotted several graphs and have worked out what size component of the main force friction was using formulas. I attempted to work out the coefficient of friction but the value that I got was out by 0.024ms-2.

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