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# 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.

Extracts from this document...

Introduction

Data Analysis Coursework

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

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)

## 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 1

The experiment was performed for eight different range’s 20cm – 160cm, increasing in steps of 20cm with 3 attempts at each distance.

Plan of calculations

Middle

1.84

1.673

1.84

1.84

1.83

1.837

## Fig 5

 Average Acceleration 0 1.929 1.691 1.490 1.287 1.375 1.206 1.154 1.098

As 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-Fr

By rearranging this formula to get the Frictional Force Fr.

A Fr = Mg-Ma

So 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 A

Fr = (0.6678 x 9.8 x cos82.3) – 1.15 x 0.6678

Fr = 0.876 – 0.768

Fr = 0.108

2.  Into formula A

Fr = (0.6678 x 9.8 x cos82.3) – 0.2 x 0.6678

Fr = 0.876 – 0.134

Fr = 0.742

3.  Into formula A

Fr = (0.6678 x 9.8 x cos82.3) –0.08 x 0.6678

Fr = 0.876 – 0.053

Fr = 0.823

The 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 graph

Fig 7

a = ΔV/ΔT

For 20cm

a = 0.72 – 0/0.373 – 0

a = 1.929 ms-2

For 60 cm

a = 1.227/0.823

a = 1.490 ms-2

For 100 cm

a = 1.540/1.120

a = 1.375 ms-2

For 140 cm

a = 1.727/1.497

a = 1.154 ms-2

From this data I will calculate the expected friction using the formula: -

Ma = Mg-Fr

Fr = Mg – Ma

 Av Acceleration Mass Gravity Friction 0 0 0 0 1.929 0.6678 9.8 5.257 1.691 0.6678 9.8 5.415 1.490 0.6678 9.8 5.549 1.287 0.6678 9.8 5.685 1.375 0.6678 9.8 5.626 1.206 0.6678 9.8 5.739 1.154 0.6678 9.8 5.774 1.098 0.6678 9.8 5.811

Conclusion

 Av Velocity Mass 0.720 0.6678

K = ½mv2

= ½ x 0.6678 x (0.72)2

= 0.173 J

The rest of the data follows suit: -

 kinetic energy (J) Av Velocity Mass 0 0 0 0.173 0.720 0.6678 0.347 1.020 0.6678 0.502 1.227 0.6678 0.645 1.390 0.6678 0.792 1.540 0.6678 0.894 1.637 0.6678 0.995 1.727 0.6678 1.126 1.837 0.6678

Next I will work out gravitational potential energy

ΔGPE = MgΔh

To work out the height at each of the distance values I will use trigonometry.

Sinθ = a/h

Sin7.7o = a/0.2

a = 0.2 x sin7.7

a = 0.03

I shall use the distance values from the following points.

Here is a sample calculation. For a ramp length of 0.2 and height 0.03

GPE = 0.6544 x 9.8 x 0.3

= 1.309

 Distance (M) Mass (Kg) Gravity GPE 0 0 0 0 0.2 0.6544 9.8 1.309 0.4 0.6544 9.8 2.618 0.6 0.6544 9.8 3.927 0.8 0.6544 9.8 5.236 1 0.6544 9.8 6.544 1.2 0.6544 9.8 7.853 1.4 0.6544 9.8 9.162 1.8 0.6544 9.8 10.471

Comparing 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: -

 GPE Kinetic Energy 0 0 1.283 0.173 2.565 0.347 3.848 0.502 5.130 0.645 6.413 0.792 7.696 0.894 8.978 0.995 10.261 1.126

Conclusion

In 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.

This student written piece of work is one of many that can be found in our GCSE Forces and Motion section.

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