Lucas Morgan

Slide acceleration practical

Contents: Introduction: 2, 3 Background: 3, 4 Part A: 5 Part B:

Practical Group: Darcy Kellock, Lucas Morgan

By: Lucas Morgan

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Lucas Morgan

Introduction:

The overall aim for our practical investigation is to investigate how the acceleration of a body will act on a slide depending on the angle of the slide (Part B) and mass of the particular body (Part A). We aim to get consistent results by trying to minimise/ control variables such as friction and reaction time. We are also commenting on the thresh hold angle for a set mass in order to find when the magnitude of friction is overcome by the force down the plain of the body.

Equipment: To investigate this we devised an experiment using the fallowing equipment:

Wooden plank approximately 2.0m long (as slide) Stop watch ( to measure the time for the object to travel down the slide) Different weights ( to change the mass of the body on the slide) Plum bob ( to accurately measure the aspects of the slide) Tape measure ( to measure the distances) Plastic container ( to contain the weights, plastic because its relatively frictionless on

wood) Plastic bag ( to contain weights in the container) Tables, chairs, boxes (to rest the slide on and provide the different angles) Sticky tape ( to use as a marker) Electronic Scales ( to weigh the plastic container, you must add this to the overall

mass of the body) Calculator (for calculations)

Diagram:

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Modifications:During the experiment we made a number of modifications, these modifications were made in order to better conduct the experiment. We originally had a sled instead of a container but we found that this sled didn’t hold the weights and it was not consistent enough with its path, so instead of the sled we decided to use the plastic container. When changing one of the masses we put in 0.675 kg’s instead of 0.75 kg’s but luckily we did the same with all of the tests. We also aimed to use a wide variety of angles but found that we couldn’t go below a certain angle because the force due to friction was too much for the container to overcome, so we only used angles above 23.5o.

Sources of error:Because we were using a wooden plank as the slide there were different areas of friction on the board, and the same with the plastic container. In some cases the container came half on and half of the slide, this would have changed the magnitude of friction on the container altering its acceleration. When this occurred we re-did that test in order to keep the conditions consistent. Because the times were recorded by a person on a stop watch, that person has a certain reaction time so the times have a tolerance of + or - 0.2 seconds. Because we measured distances using a tape measure there is a tolerance of + or – 0.001 meters.

Background Physics:

SI units: SI units are the units in which the desired magnitudes are measured. In order for the equations to work the magnitudes need to be in the right unit.

SI units/ Symbols Compnent Unit Symbol

Time Seconds (s) tDistance Meters (m) xAcceleration Meters per second2 (ms-2) aForce Newtons (N) FAngle Degrees (xo) θFinal Velocity Meters per second (ms-1) vInitial Velocity Meters per second (ms-1) uMass Kilograms (kg) m

To calculate the angle (θ) of the slide I will be using: Tan-1(Xv/Xh)

Eg: Tan-1(0.82/1.376) = 30.79o Xv

θ Xh

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Acceleration: In this practical I will be comparing a=gsinθ to calculate the acceleration of the body down the frictionless slide. a=gsinθ, where g is the acceleration due to gravity and θ is the angle that the incline makes to the horizontal. Eg: a = 9.8sin60, a= 8.5 ms-2

I will be comparing this method of calculating the acceleration to the method of a=2x/t2, where x is distance in meters and t is time in seconds. Eg: a= 2(3)/ 1.22, a=4.2ms-1

We use these two equations to give us the theoretical acceleration as well as the actual acceleration and from this we can see how much velocity is lost due to friction.

Velocity: The final velocity of a body is given by v=u+at,

Net force: or ∑F is given by the equation ∑F=mass x acceleration. This is Newton’s second law of motion, stating that the acceleration on a body experiencing an unbalanced force is directly promotional to the new force and inversely proportional to the mass of the body.To calculate the Normal force of Fn we use the equation Fn= Fgcosθ. Where Fg is mass x force by gravity (9.8)And therefore Ff=µk x Fn

Percentage difference: When determining the acceleration of a body using two different methods there is not one excepted true value of the acceleration. In these cases we use the percentage difference equation. This divides the absolute difference of the results by the average of the results:

Xr: In order to calculate the acceleration of the body we must calculate the distance it has travelled, this is Xr or X resultant. It is obtained by Pythagoras’s theorem a2+b2=c2

XrSo:Xh2+Xv2=Xr2 Xv

Xh

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where µk is the Coefficient of kinetic friction of plastic on wood, which is 0.2

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Part A: The mass of the body.

Method:1. Place slide onto either table/chair/box.2. Mark the starting point on the ramp with sticky tape and then stick plum bob to that starting

point as shown on the diagram on page 2. 3. Measure the Xv and Xh as shown in the diagram on page 2, when adjusting the height of the

plum bob the metal end should be just above the ground and stationary.4. Place desired weights into the plastic bag and then into the plastic container.5. One person on the stop watch, times how long it takes for the container to go down the

slide. The end point is when the container hits the floor.6. Record all reading and distances, and repeat steps 4 and 5 using different weights

Results: Set angle and different masses, all weights in kilograms and all times in seconds. The set angle is 30.79o

Mass 1 Mass 2 Mass 3 Mass 4 Mass 5 Mass 6 Mass 7Time (1) 1.35 1.69 1.57 1.47 1.59 1.63 1.47Time (2) 1.31 1.4 1.47 1.41 1.56 1.78 1.47Time (3) 1.41 1.63 1.5 1.4 1.54 1.75 1.62Time (4) 1.46 1.66 1.59 1.47 1.59 1.78 1.59Time (5) 1.5 1.9 1.66 1.5 1.71 1.37 1.22Time (6) 1.34 1.82 1.57 1.6 1.72 1.66 1.75Time (7) 1.56 1.53 1.57 1.78 1.47 1.47 1.63Time (8) 1.53 1.47 1.57 1.59 1.84 1.5 1.54Avg Time 1.43 1.64 1.56 1.53 1.41 1.62 1.54

1.602m0.82m

30.79o

1.376m

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# Mass (kg)"+" sled mass (0.1kg)

1 0.5 0.62 0.675 0.7753 1 1.14 1.25 1.35

5 1.5 1.66 1.75 1.857 2 2.1

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Calculations:Step1. By using the method of a=gsinθ, the theoretical acceleration of the body down a frictionless slide is:

a= 9.8 sin 30.790 a= 5.02 ms-2

Step2. By using the method of a=2x/t2, the actual acceleration down the slide was:1. Mass 1: a=2(1.602)/(1.432),

a=1.57 ms-2

2. Mass 2: a=2(1.602)/(1.642),a=1.19 ms-2

3. Mass 3: a=2(1.602)/(1.562),a=1.32 ms-2

4. Mass 4: a=2(1.602)/(1.532),a=1.37 ms-2

5. Mass 5: a=2(1.602)/(1.412),a=1.61 ms-2

6. Mass 6: a=2(1.602)/(1.622),a=1.22 ms-2

7. Mass 7: a=2(1.602)/(1.542),a=1.35 ms-2

Step3. From this we can see that quite a lot of acceleration is lost due to friction, from the percentage difference equation:

1. [(5.02-1.57)/(5.02+1.57)/2]x100 = 105% 2. [(5.02-1.19)/(5.02+1.19)/2]x100 = 123% 3. [(5.02-1.32)/(5.02+1.32)/2]x100 = 117% 4. [(5.02-1.37)/(5.02+1.37)/2]x100 = 114% 5. [(5.02-1.61)/(5.02+1.61)/2]x100 = 103% 6. [(5.02-1.22)/(5.02+1.22)/2]x100 = 122% 7. [(5.02-1.35)/(5.02+1.35)/2]x100 = 115%

Step4. To calculate the final velocity of the body we use the equation v=u+at, but because all bodies were stationary at the start u=0, so the equation becomes v=at:

1. 1.57 x 1.43= 2.24ms-1

2. 1.19 x 1.64= 1.95ms-1

3. 1.32 x 1.56= 2.06ms-1

4. 1.37 x 1.53= 2.09ms-1

5. 1.61 x 1.41= 2.27ms1

6. 1.22 x 1.62= 1.97ms-1

7. 1.35 x 1.54= 2.08ms-1

Step5. The normal force by Fn=Fgcosθ:1. (0.6x9.8)cos30.79= 5.0N perpendicular to the plain2. (0.775x9.8)cos30.79= 6.5N perpendicular to the plain3. (1.1x9.8)cos30.79= 9.2N perpendicular to the plain4. (1.35x9.8)cos30.79= 11N perpendicular to the plain5. (1.6x9.8)cos30.79= 13N perpendicular to the plain6. (1.85x9.8)cos30.79= 15N perpendicular to the plain7. (2.1x9.8)cos30.79= 18N perpendicular to the plain

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Step6. Force of friction, given by the equation Ff=µk x Fn, where µk=0.21. 0.2 x 5.0= 1N up the plain2. 0.2 x 6.3= 1.2N up the plain3. 0.2 x 9.2= 1.8N up the plain4. 0.2 x 11= 2.2N up the plain5. 0.2 x 13= 2.6N up the plain6. 0.2 x 15= 3.0N up the plain7. 0.2 x 18= 3.6N up the plain

Discussion:

As seen on page 7 the two different equations for calculating the acceleration of the body were quite different. The main reason for their difference is that the first method (a=gsinθ) does not take into account the force of friction (Ff) while the second equation (a=2x/t2) does through the factor of time. This difference is accounted for in step3 of the calculations and is shown to be quite extreme with all the results being over a 100% percentage difference.

So why is there such a large difference? It is because of the friction between the slide and the container, this is calculated in steps 5 and 6. And is represented in the table below, as shown there is a direct connection between the force of friction and the normal force. That is that the higher the normal force the higher the force of friction.

Normal Force vs Friction Force graph

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4 6 8 10 12 14 16 18 200

0.5

1

1.5

2

2.5

3

3.5

4

Normal Force (N)

Friction Force (N)

Lucas Morgan

As seen in Step2 the acceleration of the different masses were very similar to each other. As a result of this similarity and the similarity in their average times down the plain we can draw to the conclusion that if you were on a slide in real life it would make a slight but not noticeable difference if you had a higher mass or a lower mass. And this is reflected in the graph below of the bodies, this graph shows there is no connection between your mass and the acceleration you experience. So if you wanted to go faster on a slide you’re better off trying to find something with less friction to put between you and slide than to change your mass.

Acceleration vs mass graph

Part B: The angle of inclination of the slide

Method:1. Place slide onto either table/chair/box.2. Mark the starting point on the ramp with sticky tape and then stick plum bob to that starting

point as shown on the diagram on page 2. 3. Measure the Xv and Xh as shown in the diagram on page 24. Place the desired set mass into the plastic bag in our case it was 0.775kgs and then into the

plastic container.5. One person on the stop watch, times how long it takes for the container to go down the

slide. The end point is when the container hits the floor.6. Record all reading and distances, and repeat steps 1, 3 and 5 using different tables, boxes or

chairs to get the desired angles. You may need to adjust the string on the plum bob in order to measure Xv and Xh, when adjusting the steel end of the plum bob should be just sitting above the ground and stationary.

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0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Mass (kg)

Acceleration (ms

-2)

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Results: Set mass of 0.775kg

Angle 1 Angle 2 Angle 3 Angle 4 Angle 5 Angle 6 Angle 7T (1) 1.57 2.09 0.9 0.66 0.75 1.53 3.03T (2) 1.47 2.75 0.84 0.82 0.72 1.52 1.94T (3) 1.5 1.97 1.1 0.62 0.84 1.75 3.75T (4) 1.59 2.22 0.9 0.68 0.91 1.84 5.19T (5) 1.66 2.8 1 0.87 1.06 1.65 7.22T (6) 1.57 2.2 1 0.79 0.75 1.75 3.32T (7) 1.57 2.4 1.03 0.87 0.85 1.78 2.56T (8) 1.57 2.72 0.94 0.85 0.82 1.84 2.4Avg T 1.5 2.4 0.96 0.77 0.83 1.7 3.7

#Angle (º) Xv (m) Xh (m)

1 30.79 0.82 1.376

2 25.61 0.695 1.45

3 36.46 0.951 1.287

4 43.28 1.1 1.168

5 40.67 1.045 1.216

6 26.9 0.729 1.436

7 23.5 0.645 1.482

Calculations:

Angle using the equation Tan-1(Xv/Xh):

1: 2:

θ θ

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1.376m 1.45m

0.82m0.695m

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3: 4:

θ θ

5: 6:

θ θ

7:

θ

1. Tan-1(0.82/1.376) = 30.79O

2. Tan-1(0.695/1.45) = 25.61O

3. Tan-1(0.951/1.287) = 36.46O

4. Tan-1(1.1/1.168) = 43.28O

5. Tan-1(1.045/1.216) = 40.67O

6. Tan-1(0.729/1.436) = 26.9O

7. Tan-1(0.645/1.482) = 23.5O

Acceleration on a friction less slide using a=gsinθ:1. 9.8 sin 30.79= 5.0ms-2

2. 9.8 sin 25.61= 4.2ms-2

3. 9.8 sin 36.47= 5.8ms-2

4. 9.8 sin 43.28= 6.7ms-2

5. 9.8 sin 40.67= 6.4ms-2

6. 9.8 sin 26.9= 4.4ms-2

7. 9.8 sin 23.5= 3.9ms-2

Xr, but because it was a set start and end point the Xr will be the same for all angles: All angles: eg, 0.822+1.3762=2.56

Square root (2.56) = 1.6m

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1.482m

1.216m 1.436

1.168 m

0.645m

1.1m

1.045m

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Acceleration using the method of a=2x/t2, the actual acceleration down the slide was:

1. (2 x 1.6)/ 1.52= 1.4ms-2

2. (2 x 1.6)/ 2.42= 0.55ms-2

3. (2 x 1.6)/ 0.962= 3.5ms-2

4. (2 x 1.6)/ 0.772= 5.4ms-2

5. (2 x 1.6)/ 0.832= 4.6ms-2

6. (2 x 1.6)/ 1.72= 1.1ms-2

7. (2 x 1.6)/ 3.72= 0.23ms-2

From this we can see that quite a lot of acceleration is lost due to friction, from the percentage difference equation :

1. [(5.0-1.4)/(5.0+1.4)/2]x100 = 112% 2. [(4.2-0.55)/(4.2+0.55)/2]x100 = 153% 3. [(5.8-3.5)/(5.8+3.5)/2]x100 = 49% 4. [(6.7-5.4)/(6.7+5.4)/2]x100 = 21% 5. [(6.4-4.6)/(6.4+4.6)/2]x100 = 32% 6. [(4.4-1.1)/(4.4+1.1)/2]x100 = 120% 7. [(3.9-0.23)/(3.9+0.23)/2]x100 = 177%

To calculate the final velocity of the body we use the equation v=u+at, but because all bodies were stationary at the start u=0, so the equation becomes v=at:

1. 1.4 x 1.5=2.1 ms-1

2. 0.55 x 2.4=1.3 ms-1

3. 3.5 x 0.96=3.3 ms-1

4. 5.4 x 0.77=4.1 ms-1

5. 4.6 x 0.83=3.8 ms-1

6. 1.1 x 1.7=1.8 ms-1

7. 0.23 x 3.7=0.85 ms-1

The normal force by Fn=Fgcosθ:1. (0.775x9.8)cos 30.79 =6.5N perpendicular to the plain2. (0.775x9.8)cos 25.61 =6.8N perpendicular to the plain3. (0.775x9.8)cos 36.46 =6.1N perpendicular to the plain4. (0.775x9.8)cos 43.28 =5.5N perpendicular to the plain5. (0.775x9.8)cos 40.67 =5.7N perpendicular to the plain6. (0.775x9.8)cos 26.9 =6.7N perpendicular to the plain7. (0.775x9.8)cos 23.5 =6.9N perpendicular to the plain

Force of friction, given by the equation Ff=µk x Fn, where µk=0.21. 0.2 x 6.5=1.3N up the plain2. 0.2 x 6.8=1.3N up the plain3. 0.2 x 6.1=1.2N up the plain4. 0.2 x 5.5=1.1N up the plain5. 0.2 x 5.7=1.1N up the plain6. 0.2 x 6.7=1.3N up the plain7. 0.2 x 6.9=1.4N up the plain

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Discussion:

As expected in the results there are noticeable changes in the body’s acceleration as the angles changes. This is to be expected because there was no change in the body’s mass so the friction force should be relatively constant, which is show in the results. So with friction being constant the only thing effecting the body’s acceleration is the angle of the slide. And this is shown in the graph below. There is a clear connection between the acceleration of a body and the angle of the slide.

If friction was ignored the graph would still have the same trend.

As mentioned before the force of friction has remained relatively constant, this is because the mass of the body hasn’t changed.

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20 25 30 35 40 45012345678

Acceleration(not incl friction) vs Angle

Acceleration (ms

-2)

Angle (o)

20 25 30 35 40 450

1

2

3

4

5

6

Acceleration (inc friction) vs Angle