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Friction, Work, And Kinetic Energy - Lab Report Example

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Trying to slide any object leads to creation of an opposite force , that demands more force to be applied to the object for it to move hence more work is required. The opposite force that opposes the force being applied to the box is static friction. Application of a light…
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Friction, Work, And Kinetic Energy
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INTRODUCTION Trying to slide any object leads to creation of an opposite force , that demands more force to be applied to the object for it to move hence more work is required. The opposite force that opposes the force being applied to the box is static friction. Application of a light horizontal push that does not move the box reduces the static friction force which is also directly opposite to the push. Pushing it harder increases the frictional force so as to match the magnitude of the push. There is a limit to the magnitude of static friction, that eventually makes it possible to apply a force larger than the maximum static force, making the box to start moving.
Static friction is denoted as Fstatic. An inequality has been modified; Fstatic ≤ ms N where i ms is the coefficient of static friction and N is the normal force exerted by a surface on the object. The normal force is defined as the perpendicular component of the force exerted by the surface. In this experiment, the normal force is equal to the weight of the object because it’s on a horizontal surface
Exertion of enough force to move the box requires that the same or more force be applied to keep the box moving to the desired direction to prevent loss of equilibrium which will cause it to stop moving. The friction acting on the box while it is moving is called kinetic friction (Hugh and Roger 2008). In order to slide the box with a constant velocity, a force equivalent to the force of kinetic friction must be applied. Static and kinetic friction depends on the surfaces of the box and the floor and on how hard the box and floor is pressed together. Kinetic friction is modeled with Fkinetic = µk N where µk is the coefficient of kinetic friction. In general, this coefficient is less than that for static friction.
OJECTIVES:
1) To investigate the effect of the force of friction on the motion of an object using a motion detector
2) To determine the acceleration of the object, its velocity, and the distance it travels when the force of friction is applied to the object.
3) To calculate:
the force of friction and the coefficient of kinetic friction
the work done by the force of friction for different values of the object’s displacement, and the change of object’s kinetic energy.
4) To verify the work – kinetic energy theorem
APPARATUS
Computer
Vernier Motion Detector
Container with a plastic bottom surface
Electronic scale
Mass set
Track
TASKS
part 1
The mass of the container was measured and recorded. Motion detector was connected to the PC with a USB cable. Kinetic friction icon on the desktop was double clicked to open the program for data collection and analysis
A plastic ball was placed on the track and the track leveled by adjusting the track’s feet for the ball to stay in rest. The Motion Detector was placed on the track and positioned so that it could detect the motion of the block as it was sliding towards the detector.
sliding the block towards the motion detector such that the block left the hands to slide o a stop was practiced. Rotation of the block was minimized.
The block was supposed to slide for 0.3-0.5m before stopping and it was not supposed to go closer to the motion detector for more than 0.2m.
Taking data
Collection of data was done by clicking .

A total of five measurements for the empty blocks were collected.
masses totaling 1000 g were placed in the container. Results on difference in acceleration values were recorded in the data table.
Data tables :
Containier
mass, kg
error in mass, kg
w/out add mass
0.0931
w/ 1000 g
1.088
Table I: data and calculations of the container without additional mass
trial
acceleration m/s^2
N (kinetic friction force)
µk
1
2.425
0.2257675
0.24744898
2
2.399
0.2233469
0.244795918
3
2.384
0.2219504
0.243265306
4
2.342
0.2180402
0.238979592
5
2.292
0.2133852
0.233877551
Average
= 0.22049804
=0.005330888
Table II: data and calculations of the container with 1000g additional mass
Trail
Acceleration (m/s2)
Kinetic friction force ()
k
1
1.917
2.085696
0.195612245
2
1.827
1.987776
0.195612245
3
1.843
2.005184
0.195612245
4
1.965
2.13792
0.195612245
5
2.016
2.193408
0.195612245
Average
=2.0819968
=0.008166963
QUESTIONS:
Q-1 For container with additional mass For container without additional mass
Kinetic friction force for each trial: Kinetic friction force for each trial

Trial 1- 2.085696 Trial 1 - 0.24744898
Trial 2-1.987776 Trial 2-0.2233469
Trial 3-2.005184 Trial 3-0.2219504
Trial 4-2.13792 Trial 4-0.2180402
Trial 5-2.193408 Trial 5-0.2133852-
Coefficient of kinetic friction for each trial: Coefficient of kinetic friction for each trial
Trial 1-0.195612245 Trial 1-0.24744898
Trial 2-0.195612245 Trial 2-0.244795918
Trial 3-0.195612245 Trial 3-0.243265306
Trial 4 -0.195612245 Trial 4-0.238979592
Trial 5-0.195612245 Trial 5-0.233877551
Average for
For container with additional mass= 0.008166963
For container without additional mass=0.005330888
Standard deviation in the coefficient
Container without additional mass=0
Container with additional 1000g mass=0.005330888
Error in the average of the coefficient:
Container without additional mass=0.00004653782
Container with an additional 1000g mass=0.0000567438
Q –2
Does the coefficient of kinetic friction depend on speed?
No, the kinetic friction does not depend on speed. It only depends on the surfaces in contact. From the data collected during the experiment, it is evident that the speed/velocity of the sliding block does not affect the coefficient of the kinetic friction.
Q—3
Does the force (not coefficient) of kinetic friction depend on the weight of the sliding block? Explain.
No, the force of the kinetic friction is independent of the weight of the sliding block. This is because the weight of the block is balanced by the normal force.
Q—4
Does the coefficient (not force) of kinetic friction depend on the weight of the sliding block? Explain.
No, the coefficient of the kinetic friction is independent of the weight of the sliding block because it only depend on the nature of the two surfaces in contact and the area.
part ii
“Collect” button was clicked and the block moved to slide toward the Motion Detector, so as to collect data on position and velocity of the block. This was repeated Repeat several times. Velocity vs time graph was observed a part of the linear change in velocity was selected by holding down the left button moving across the graph. Value of x was noted. Values as and were recorded in the Data Table III. Values of velocity corresponding to these x values were also recorded.
Table III - position and velocity data
data set #
x1, m
v1, m/s
x2, m
v2, m/s
Qv
run 1. no add mass
0.69842
-0.76831
0.581952
-0.13788
0.027283
run 2 mass added
0.69468
-0.6607
0.589551
-0.1575
0.019188
Error in velocity was determined and recorded.
The force of friction was the only force acting on the block in the direction parallel to the block’s displacement, so the net force is the force of friction:
F = Ffr= kFN was calculated.
Its error was too calculated using formula:

Value of F and its error were recorded in Table IV.
Change in KE and its error was Calculated as KE = and recorded in table iv. Because the error in the mass is much less than the error in the velocity, the error in was given by:
(Hugh and Roger 2008)

Verification of work—kinetic energy theorem:
W = KE. This means the difference should be zero. Difference was calculated and recorded in table iv using formular:

The error in this difference was given by formular below and recorded into the table
Table IV – calculations of work and change in KE.
Data set #
F
N
N
x2-x1
m
W
J
J
J
J
diff = W-DKE
J
sdiff
J
Run 1 (no mass)
0.201178007
0.002175156
-0.116468
-0.0234308
-0.000253336
-0.02659352
0.00198272
0.00316272
0.001998839
Run 2 (added mass)
2.265212518
0.038943105
-0.105129
-0.238139527
-0.00409405
-0.223974723
0.014179627
-0.014164804
0.01475883
CONCLUSION:
The theorem came out to be true because the statistical differences between work and kinetic energy were within the statistical errors from zero. The difference i.e. W-DKE from the container with no additional mass was 0.001998839 while W-DKE from the container with additional mass was 0.01475883. Hence they were within the statistical errors of each other . so the results from the experiment turned out as expected.
WORKS CITED
Hugh D Young and Roger A. Freedman .University Physics (12th edition), Addison-Wesley CO LTD, USA, page 329. 2008 .Print Read More
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