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.

I will use the equations F = Ma and a = ΔV/ΔT.

Expected Values

Fig 2

To find velocity A I will break up the downward force C. I know the angle of the ramp is

Sin  = o/h. =0.327/2.44 = 0.134. Sin-1   = 7.701o

The other angle (B) in the triangle = 180-7.701-90 = 82.3o

So breaking up the force gives us: -

                                                  Fig 3

Original 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 = 0.877

And 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-2

The 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 analysis

Data that was gained from the experiment: -

To calculate the average acceleration, I used a = ΔV/ΔT

Fig 5

        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

From 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 friction

Now 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 Fup

The friction data was taken from fig 6.

Inserting the data for gives me

0.108 = coeff Fr x (6.544 x cos84.2)

Rearranging gives me an answer of: -

0.108/0.865 = 0.125

By subtracting this value from the perfect results I should find the actual force of acceleration.

0.877-0.125 = 0.752

Now by using a = F/M I should find the acceleration that I found in part 1).

a = 0.752/0.6678

a = 1.126 ms-2 

This 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 = ½mv2

        This 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.

ΔGPE = MgΔh

        To start I will calculate the kinetic energy.

Starting with the following data I will perform a sample calculation.

K = ½mv2

= ½ x 0.6678 x (0.72)2

= 0.173 J

The rest of the data follows suit: -

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

        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: -

        

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.