AP Physics Lab 6

Gabrielle Murphy (insertion of graphs, objective, creation and insertion of data table, deformation vs. height in conclusion, quantitative models in conclusion, force vs. spring deformation in conclusion), Ryan Partain (creation of graphs, sources of error, beginning of deformation vs. speed in conclusion, force vs. spring deformation in conclusion), Trey Seabrooke (procedure), Tyler Kolby

Objective:

To determine the mathematical and graphical relationship between the distance the spring is compressed (deformation of the spring) and the force of the spring, the maximum velocity, and the maximum height.

Procedure:

1. Set up an apparatus including: a track(angled 1.2° up from parallel to the ground), a cart, a spring, a force probe attached to the spring, and a motion detector.

2. Place the spring where the track forms the angle with the horizontal. Attach the force probe to the spring.

3. Carefully rest the cart on the spring, with 0 compression of the spring. This is the starting position.

4. Start the motion detector.

5. Compress the spring by pushing the cart against it.

6. Release the cart and record data gathered by the force probe and the motion detector.

7. Repeat steps 3-6 with decreasing compression (less force applied to cart) of the spring each successive time until enough data is collected for the experiment to be valid.

Data Table:

Figure 1

Graphs:

Force vs Deformation (Figure 2)

Position (r)vs Time (Figure 3)

Time (s)

Velocity vs Time (Figure 4)

Height vs Deformation (Figure 5)

Height vs Area (Figure 6)

Velocity vs Deformation (Figure 7)

Velocity^2 vs Area (Figure 8)

Conclusion:

DEFORMATION VS FORCE

In this experiment, the force exerted on the spring was measured using a force probe attached to the spring. The force required to compress the spring, measured by the force probe, is directly proportional to the distance the spring was compressed, x.  In other terms, the force and spring deformation are directly proportional (as force increases, the spring deformation increases in the opposite direction). Because force and spring deformation are directly proportional, the graph is linear (Fig. 2) and it is in Quadrant II. The slope can be found by the force divided by the spring deformation (N/m). The slope of this graph, represented by the letter k, is the spring constant. The equation created by the force vs spring deformation graph is F=k(x) where k represents the “stiffness” of the spring, or how much the spring resists deformation by a force exerted on it.

The area of the graph can found by multiplying F and x. Because the graph makes a triangle, the area of a triangle equation can be used:

A=(½)bh

where b is substituted by xand h is substituted by F. It can be represented by the equation:

A=(½)xF

Because F= k(x), F can be substituted by k(x) in the area equation. The new equation becomes

A=(½)k(x)^2

This area represents the force that the spring exerts on the cart (the system) that causes the displacement of the cart, which is the work done on the system by the spring. When the system does not include the spring, the equation is

W=(½)k(x)^2

In a system that includes the spring, the energy in the spring is called elastic energy, or U(el). The general equation becomes

U(el)= (½)k(x)^2

DEFORMATION VS HEIGHT

The Position vs Time graph (Fig. 3) can be used to find the relationship between spring deformation (dependent variable) and the maximum height the cart reached as a result (independent variable). Because the track was at an angle, the distance the cart travels on the track, r, can be broken down into two components: xand y where x is the horizontal distance and y is the vertical distance. x, the change in the horizontal, is also the deformation of the spring because it is the horizontal distance the cart and the spring moved when the spring was compressed. x can be found using the minimum value of the Position vs Time graph (Fig. 3) since the cart moves backwards and reaches its minimum position when the spring is at its maximum compression. r is the maximum value of the graph (Fig. 3) because it is the greatest distance traveled by the cart on the track and where it reaches its maximum position. Because the track formed an angle with the horizontal, y, the maximum height, can be found using trigonometry:

h= r(sin)

where =1.2° and r is the maximum of the Position vs Time graph:

h(max)=(0.242 m)(sin 1.2°)= 0.0051 m

The values for h found using the general equation h=r(sin 1.2°) can be graphed in terms of spring deformation,  x to create the Height vs Deformation graph. This would produce a parabolic graphical relationship, as seen in the Height vs Deformation graph (Figure 5). The graph can be linearized by squaring the x-axis - x. The equation for this relationship is

h=NUMBERS(x)^2

The slope of the linearized graph of deformation vs height does not shed light on its relationship since there is no constant in the experiment that can be applied to the slope. Looking at the linearized graph, the height is quadrupled because xis squared. This can be compared to the Force vs Deformation graph (Fig. 2), since xis also a variable in that graph. When xis squared in the Force vs Deformation graph (Fig.2), force becomes squared (because of its linear relationship) and the area under that graph is quadrupled. The two variables affected by the squaring of xthe same way are then be compared to each other to see if there is a meaningful relationship between them: height and the area under the Force vs Deformation graph, or the work done on the system.

To compare Height vs Work,

This new relationship shows that deformation of the spring, which has a direct relationship with how much work the spring does on the cart, will affect the maximum height of the cart.

DEFORMATION VS SPEED

The Velocity vs Time graph (Fig. 3) can be used to find the relationship between spring deformation and the speed the cart traveled during the experiment.  The initial negative velocity on the graph represents when the car was being pushed into the spring, compressing it.  The sudden jump in the velocity is when the car was released from the spring, and it quickly reached its maximum velocity.  Since the track was at a 1.2° angle, the cart slowed as it progressed up the ramp.  The cart eventually reached a velocity of 0 m/s and then started traveling backwards with a negative velocity.  In this experiment, the force of the spring decreased with each trial.  This is the dependent variable, and it caused the velocity to change as a result.  This causes the velocity to be the independent variable.  These results were impacted by the shape of the ramp as well.  Because the ramp was tilted at a 1.2° angle, it caused the cart to slow as time progressed.  The acceleration was in the negative direction because of the Earth’s gravity.  The forces acting on the cart were unbalanced, and gravity was the strongest, pulling the cart down.  This caused the velocity to decrease, reach zero, and then increase in the negative direction as time progressed.

The equation derived from this graph (Fig. 3) is V(max)=

EQUATION MODELS

W=Fx

U(el)=(½)k(x)^2

U(g)=mgh

U(k)=(½)mv^2

SOURCES OF ERROR

During this experiment, many machines were used, and it relied less on human actions.  This reduces the possibilities of errors in the experiment, but there are still possible sources present.  The spring could have been compressed too much or not enough during the experiment.  This would cause inaccurate readings on the force probe, which would lead to errors throughout the experiment.  Since the distance the cart traveled was dependent on the distance the spring was compressed, this reading would also be inaccurate.  Finally, the different amounts of forces applied to the spring could have been inconsistent.  The force applied to the spring was supposed to decrease steadily and constantly, and it may not have been decreased constantly.  This would be yet another source of error during the experiment.

AP Physics Lab 5

Tyler Kolby, Gabrielle Murphy, Ryan Partain, Trey Seabrooke 

Friction Lab

Objective:
To determine the graphical and mathematical relationship between normal force, surface texture, and surface area in max friction (static) and constant velocity friction (kinetic)

Apparatus:

Procedures:

1. Measure the mass of the block.

2. Place the block on its side. Connect the spring scale to the hook on the block.

3. Pull scale slowly until block starts moving at a constant velocity, making sure to pull

   horizontally.

4. Watch the force of tension on the scale (record the maximum tension and tension at

   constant velocity).

5. Add weights to the block 10 times, adding about 50g each time.

   6. Repeat steps 14 for each new trial, starting at the same position. Data Table:

   
   

   
   

   Graphs:

   Maximum Force vs Normal Force Graph 

   
   

   
   

   Constant Velocity Force vs Normal Force Graph

   
   

   

   Conclusion:

   Because the block did not move vertically, the normal force the table exerts on the block is equal to the force of gravity the Earth exerts on the block, which is the block’s weight. The normal force can then be calculated by using the Field Constant, 10 N/kg, multiplied by the mass of the block in kg. As more mass was added to the block, the weight of the block also increased.

   The maximum force vs normal force graph shows a direct, linear relationship. As the normal force the table exerted on the block increased, the maximum force applied to the block increased. This graph shows the relationship between the normal force and the friction force when it was not moving, or static. The graph can be described by the equation: F(max) ≤ (0.643)F(N), where F(max) equals the force of friction.

   The block did not move until a force greater than the friction force was applied to it. At the maximum force, the force of tension exerted on the block was equal to the force of friction exerted on the block. We can say that the maximum force applied to the block was a friction force because it was the greatest amount of force that had to be overcome before the block started moving and that the force of tension was equal to the force of friction at that instant. Because the

   The constant velocity force vs the normal force graph also has a direct, linear relationship. As the normal force increased, the force at constant velocity increased. This graph shows the relationship between the normal force and the friction force when the block was moving, or kinetic, and the slope of this graph is smaller than the slope of the graph when the object was static. It can be described by the equation: F(CV)=(0.420)F(N).

   Normal force is compared to the friction force instead of mass because the normal force of the block is a measure of the relationship between the surface of the block and the surface of the table (the magnitude each object “pushes” the other). The friction force also measures the relationship between the surface of the block and the surface of the table (the magnitude each object “slides” against the other). Mass, on the other hand, is the measure of how much matter is in an object. This measurement would not change from surface to surface or place to place (planets with different gravitational pulls). The normal force and the friction force can depend on these changes, which affect the way two objects relate to each other.

   When the experiment was done on a different surface (a smoother floor), the slope of the graphs were smaller than the slope of the graphs on the table surface. There was a smaller force of friction between the floor and the block compared to the force of friction between the table and the block; however, the relationship between the friction force and the normal force could still be described using the same general equation.

   The overall general equation from this experiment is: F(F)=(μ)F(N), where F(F) is the maximum friction force a surface can exert on an object. This relates the maximum force of friction to the normal force using the coefficient of friction, μ, which can vary depending on the surface and the normal force acting on the object. The equation simply allows one to find the force of friction if given the normal force and the coefficient of friction, and vice versa.

   Like in all experiments, there is room for a source of error. We could have pulled the scale too quickly and caused it to be difficult to find an accurate reading of the force applied to the block. We also could have read inaccurately off of the scale and written down the wrong measurements. This applies to both the reading for the maximum force and the constant velocity force. The scale measuring the mass also could have been off by a little, throwing off the calculations later on. 

AP Physics Lab 4

Balanced Forces Particle Model Lab

Tyler Kolby, Gabrielle Murphy, Ryan Partain, Trey Seabrooke

Objective

To determine the mathematical and graphical relationship between the mass of an object and the force of gravity acting upon it.

Picture of Apparatus

Procedure

1. Obtain a balance, a spring scale, a stand, and 22 washers.
2. Set up the stand. Attach the spring scale onto the stand.
3. Measure the mass of two washers on a balance and record in the data table.
4. Place the washers on the metal piece suspended from the hook on the spring scale.
5. Record the force of gravity measured for the washers in the data table.
6. Repeat steps 3-5 ten times, adding two more washers to the previous number of washers and recording the measurements in the data table.

Data

Force of Gravity vs Mass Graph

Analysis and Conclusion

The force of gravity vs mass graph shows that as the masses of the washers increase, the force of gravity also increases. This is a direct, linear relationship between the force of gravity and the masses of the objects. Because of this linear relationship, the general equation y=mx+b can be used. In this experiment, y can be replaced by the force of gravity, FG. Additionally, x can be replaced by the mass, m. The y-intercept, b, would be zero - if there were no mass on the spring scale, the force of gravity acting on that mass is zero. The slope, m, can be calculated from the graph, resulting in: m=9.7N/kg. The mathematical relationship to describe this graph can then be explained through the following equation: FG=(9.7N/kg)m.

The slope of the graph, 9.7N/kg, can be related to Earth’s gravitational force on all objects - for every mass on earth, the gravitational force is the mass multiplied by the constant, 9.7N/kg. Due to errors in the experiment, this slope is slightly off, and the value should be closer to 10N/kg. The slope of the graph represents the strength of the Earth's gravitational force field - the gravitational force for each mass at that point in space. A more generalized equation that can be used to represent this would be FG=gm, where "g" is used to represent the slope and therefore the strength of Earth's gravitational force field - 10N/kg.

Some errors were present during this experiment. Each washer was a different mass; because of this, the addition of washers would not have been in equal increments. The result would be a variation in the data as the masses of the washers increased. Another source of error is that the balance used may not have been the most accurate, so the measurement of the masses could have been a little off. We also might not have had the most accurate scale to hold the washers. We could improve this if we used more accurate materials to measure and washers with more similar masses.

AP Physics Lab 3

Momentum Transfer Model Lab

Katie O'Byrne

Michael O'Connell

Chris Prattos

Ryan Partain

Objective: to determine the graphical and mathematical relationship between the ratio of masses to the ratio of velocities of two carts exploding.

Our group's ratio - 4:2

Materials and diagram:

Track with stoppers

One cart without a spring (508 g)

One cart with a spring (513 g)

4 individual weights (~500 g)

Ruler

Balance

Procedure:

1) place both carts on the track without weights

2) set off the spring with a ruler until you find the position at which they hit the opposite stoppers at the same time

3) Record your starting position (ours was 71 cm)

4) Find and record the displacement between starting and ending position for each cart and find the ratios of cart A's displacement to cart B's displacement (d1,d2)

5) Put one weight on cart A for a ratio of 2:1

6) Repeat Steps 2-4

7) Use additional weights to attain the following approximate ratios of the carts masses while(3:1, 3:2, 4:1, 4:2 and 4:3) , and repeat steps 2-4 for each

         Note~ when attaining the ratios, the masses may not be exactly 2:1, 3:1, ect, find the exact ratios of the masses and record them in the Data Table

8) find the inverses of your previous ratios so as to attain 1:2, 1:3, 2:3, etc. and record the data

Data Analysis:

Conclusion:    

    When we first graphed the data, it formed an inversely proportional relationship. As the mass of the carts increased, the displacement of the carts decreased. in order to linearize the data we took the inverse of the x-axis or mass values. This created a linear line of  which we were able to find the slope. This gave us the equation d1/d2= (0.72) M2/M1. However displacement  can be rewritten as velocity (V1/V2) for both carts traveled their distances in the same amount of time. Also, because one cart’s displacement was negative for it moved in a negative direction the equation must be V1/V2= (-0.72) M2/M1. The slope in this equation represents the amount in which the momentum of the two carts changed when the mass increased.

    In this lab, the data for each ratio came from a different group in the class. This left room for some errors such as each group’s track was a different length, thereby changing the starting and stopping points of the carts. We also could have hit the button that released the spring differently each time, either from an angle or with different force. Repeated trials could have loosened the stoppers by being repeatedly hit by the carts that could lead to slight errors in the displacement measurements.  One more possible source of error could be from slight variations in the starting points of the cars.  It looked the same to the human eye, but it is possible that the starting position changed a little each time we hit the cars.

AP Physics Lab 2

Lab #2

Katie O’Byrne, Michael O’Connell, Chris Prattos, Ryan Partain

Constant Acceleration Particle Model Lab

Objective: To determine the graphical and mathematical relationship between position and clock reading for a ball rolling down an incline.

Materials and Diagram:

-Ramp (with 10 cm increments)

-Stand with ramp attached at hole 12

-Metal Ball

-Photo Gates

-Photo Gate Timer

Procedure:

1) Set up stand and attach ramp at hole 12 from the bottom

2) Place photogate A at 10cm and photogate B at 20cm.

3) Clear all readings on the photogate timer

4)Place metal ball at top of ramp and release

5) Record time in seconds for photogate A, photogate B, and the time it took to get to photogate A to B.

6) Repeat steps 3-5 leaving photogate A at 10 cm and moving photogate b in 10 cm increments

7) On trial 8 skip 90cm and move photogate B to 100 cm for the final trial.

Data:

Table for Velocity-Time Graph

VB=(d/tB)

d=1.90cm

(For the VB and tB quantities, view the graph above)

Data Analysis:

X=(215.40cm/s^2)t^2+(80.544cm/s)t+10.637cm

X=(349.955cm/s^2)t^2+20.705m

V=(425.241cm/s^2)t+89.7cm/s

Conclusion:

1. Describe the motion studied in terms of both position and speed and explain the motion map

which represents the motion. Include qualitative motion map showing velocity and acceleration

vectors.

The position changed at faster rates per second and therefore the motion had increasing velocities. The object traveled a greater distance per second as the ball continued down the ramp. The velocity then increased as shown by the increasing length of the arrows in the motion map. The change in velocity (acceleration) however was constant.as shown by the same sized arrows in the motion map .

2. Explain the relationship between variables seen in the x-t and x-t^2 graphs. Show and explain the equations you developed and include numerical values for your constants.

As the object rolled down the ramp , it  changed positions  at a faster rate per second. This caused the graph to form a quadratic relationship or curve.  The equation for the position time graph  is x=½at^2+ V0t+ Xo.  A represents acceleration, V0 represents the velocity at time zero and X0 represents position where time is zero.  T represents time, while x represents position.  The equation for this graph was  X=(215.40cm/s^2)t^2+(80.544cm/s)t+10.637cm; however, in order to find acceleration one needs to find the slope of the velocity, which is the slope of the position vs time graph. In order to find the slope of the position vs time graph we first had to linearize the data. We did this by squaring the time values  which formed a straight line.  This gave us the equation  x= V0t^2+ X0. The equation for this graph was x= (349.955cm/s^2)t^2+20.705m.

We were able to then find the slope of this line but what it represented wasn’t clear based on the data.  We then created a velocity-time graph in order to understand the meaning of the slope.  The velocity time graph created linear line showing that the velocity was increasing steadily as the ball rolled down the ramp. The ball was accelerating at a constant rate as it rolled down the length of the ramp.  

3. Explain how you determined the velocity at a given point in order to make the v-t graph. (This

will require a bit of detail.)

We took the diameter of the ball, 1.90 cm, and divided it by the tB value. the diameter of the ball is the distance it travelled as it rolled. The time it took to travel from the start of gate b to the the end of gate b is the time elapsed)This gave us the velocity at each of the points so we could make the v-t graph.  We plotted the velocity (VB) on the y-axis and the time (tAB) on the x-axis.  The velocity was measured in cm/s, the time was measured in s, and the slope (acceleration) was measured in cm/s^2.

4. Explain the relationship between variables seen in the v-t graph and what this means in terms of the object's motion. Show and explain the equations you developed and include numerical values for your constants.

The relationship is linear (y=mx) or (V=(425.241cm/s^2)t (s))

As the object rolls down the incline the velocity of that object as it enters and exits photo gate b changes as the distance between the ball’s starting point and photogate B increases. This is because the metal ball has a constant acceleration so its velocity is constantly changing. As time goes on the velocity increases linearly as seen from the equation above.

5. Explain the significance of the slope of the v-t graph and provide an interpretation for the slope

without using the word "acceleration." Then provide a formal physics definition of acceleration.

The slope of the v-t graph shows the rate of change in the velocity per second and can theoretically be used to predict the velocity for any given second. Because the slope of the v-t graph is positive instead of zero it means that the object has an increasing change of position per elapsed time meaning that not only is the object changing positions per second but the length of this displacement is also changing.

Acceleration - The rate of change in velocity per unit of time

6. Show and describe how a general velocity equation (consisting entirely of variables) is developed from the v-t graph.

When the v-t graph has been created, an equation can be used as a mathematical model of the graph.  Just like in an x-t graph, the slope is represented by a value.  The slope in a v-t graph is the acceleration.  Instead of using the starting position (like in an x-t graph), the starting velocity is used.  The basic equation for the graph (graph of a line) is y=mx+b.  When using the units and symbols, the final equation becomes V=at+V0.

7. Show and describe how a general position equation (consisting entirely of variables) is developed from the x-t^2graph.

A position equation can be developed from the x-t^2 graph by using the slope of the x-t^2 graph to develop a v-t graph to find the area which is  ∆x which thereby must be equal to ½ of (v+v_0) t. When one makes an a-t graph, one can reason that (v+v_0)=v_0+at+v_0 therefore ∆x=½ (2v_0+at)t which can be simplified to ∆x=v_0t+½ at^2 and then x-x_0= v_0t+½ at^2 into x=½ at^2 + v_0t +x_0.

Some possible sources of error could have been that our photogates could have been imprecise and that we could have dropped the ball at slightly different areas than the exact top of the ramp due to our fingers being slightly sticky from handling our old and sticky photogates. There may also have been slight changes in the air currents in the room as the AC cut on and off but these sources of error probably gave negligible results.