Lab 6a1:
Work and Energy
Lab Course:
Physics 111A-107 – Physics 1 Lab
Teacher’s Name:
Brian E. O’Donnell
Date:
October 18, 2019
Introduction
Work is the measure of energy transfer that occurs when an object is moved over a distance by
an external force which is, fully or partly, applied in the direction of the displacement. Work is a
scalar quantity defined as 𝑊 = 𝐹⃗ ∙ 𝑠⃗ = 𝐹𝑠𝑐𝑜𝑠 and has a SI unit of ‘joule’ (J). Work can also be
expressed in terms of kinetic energy using the work energy theorem, which states that the work
done by the force on an object is equal to the change in the kinetic energy of the object. Therefore
1
1
𝑊 = ∆𝐾𝐸 = 𝐾𝐸2 − 𝐾𝐸1 . Kinetic energy is equal to 2 𝑚𝑣 2 , ∴ 𝑊 = 2 𝑚(𝑣2 2 − 𝑣1 2 ).
This lab will use a frictionless air-track to determine the work done on a glider while being pulled
by a constant force on horizontal and inclined tracks and to demonstrate the work energy
theorem. The change in the kinetic energy of the glider will also be determined. While using the
horizontal track the net force acts in the same direction as the motion of the glider, so the work
done is equal to the work done by the tension. Also, while using the inclined air track it is derived
that work is equal to the sum of the change in kinetic energy and the change in potential energy;
meaning 𝑊𝑇 = ∆𝐾𝐸 + ∆𝑃𝐸.
Objectives
1. Demonstrate the work energy by measuring the work done on and the change in kinetic
energy of an object being pulled by a constant force.
2. Gain an understanding of how the total work done on an object changes its energy.
Experimental Procedure
Apparatus/Setup
Computer with capstone software, 850 Universal interface, Force sensor, Rotary motion sensor,
Air track and supply, Glider, 50g weights(x4), Hook, String, Protractor, Wooden block, Pulley,
Right angle clamp, L-shaped rod, Table clamp.
Set up the force and rotary motion sensor using the
L-shaped rod, angle clamp and the table clamp.
Place the hook and string on the glider and place the
glider on the track. Place the string in the pulley and
use it to attach the glider to the force sensor. Place
the air track parallel to the horizontal in part 1 and
then raise the track to an angle in part 2.
Method/Procedure
1. Find the mass of the glider with the hook and weights attached and record the value in
table 1.
2. Set up the apparatus as explained above.
3. Connect the force sensor to PASPort 1 and the rotary motion sensor to PASPort 2. Turn
on the universal interface.
4. Open Lab 6 file on the computer and go to ‘Part 1: Measurements’.
5. Zero the force sensor while no force is being applied.
6. Place the glider at the starting position and old it there.
7. Press record and release the glider. The software will record the information.
8. Select 4 different point of position and velocity form the graphs. Also find the average of
the force graph. Tabulate all values.
9. Complete Table 1 using all the data that was obtained.
10. Repeat steps 6 to 9, this time with the track raised to a measured angle of inclination.
11. Complete Table 2 using all the data that was obtained
Results
Data Collected
Table 1-1
Position
Xi (m)
Vi (ms-1)
Fi (N)
1
0.051
0.336
0.43
2
0.245
0.742
0.43
3
0.519
1.074
0.43
4
0.836
1.358
0.43
Table 1-2
1
𝐾𝐸𝑖 = 𝑚𝑣𝑖 2
2
1
𝐾𝐸𝑓 = 𝑚𝑣𝑓 2
2
(J)
0.107
(J)
|𝑊 − ∆𝐾𝐸|
∙ 100
(𝑊 + ∆𝐾𝐸)/2
0.083
(J)
0.022
0.085
2.381
0.274
0.118
0.107
0.224
0.117
0.851
3→4
0.317
0.136
0.738
0.785
0.338
0.359
0.359
0.135
1→4
0.224
0.022
0.337
0.296
Position
Change
i→f
𝑠 = 𝑥𝑓 − 𝑥𝑖
𝑊 =𝐹∙𝑠
(m)
(J)
1→2
0.194
2→3
∆𝐾𝐸 = 𝐾𝐸𝑓 − 𝐾𝐸𝑖
%Difference
Table 1-2
Vi (ms-1)
0.285
Fi (N)
1
Xi (m)
0.078
2
0.215
0.479
0.43
3
0.383
0.639
0.43
4
0.567
0.776
0.43
Position
0.43
Table 1-2
Position
Change
i→f
𝑠 = 𝑥𝑓 − 𝑥𝑖
𝑊 =𝐹∙𝑠
∆𝐾𝐸 = 𝐾𝐸𝑓 − 𝐾𝐸𝑖
(m)
(J)
(J)
1→2
0.137
0.059
2→3
0.168
3→4
1→4
Calculations
∆𝑃𝐸 = 𝑀𝑔∆ℎ
∆ℎ = 𝑠 ∙ 𝑠𝑖𝑛𝜃
∆𝐸 = ∆𝐾𝐸 + ∆𝑃𝐸
%Difference
(J)
|𝑊 − ∆𝐸|
∙ 100
(𝑊 + ∆𝐸)/2
0.029
(J)
0.036
0.065
9.677
0.072
0.035
0.045
0.080
10.526
0.184
0.079
0.038
0.049
0.087
9.638
0.489
0.210
0.101
0.130
0.231
9.524

Part 1 (% Difference)
1→2
2→3
|𝑊−∆𝐾𝐸|
 %𝐷𝑖𝑓𝑓 = (𝑊+∆𝐾𝐸)/2 ∙ 100
|.083−.085|
 %𝐷𝑖𝑓𝑓 = (.118+.117)/2 ∙ 100
 %𝐷𝑖𝑓𝑓 = (𝑊+∆𝐾𝐸)/2 ∙ 100
 %𝐷𝑖𝑓𝑓 = (.083+.085)/2 ∙ 100
.002
|𝑊−∆𝐾𝐸|
|.118−.117|
.001
 %𝐷𝑖𝑓𝑓 = .084 ∙ 100
 %𝐷𝑖𝑓𝑓 = .1175 ∙ 100
 %𝐷𝑖𝑓𝑓 = 2.381%
 %𝐷𝑖𝑓𝑓 = 0.851%
3→4
1→4
|𝑊−∆𝐾𝐸|
 %𝐷𝑖𝑓𝑓 = (𝑊+∆𝐾𝐸)/2 ∙ 100
|.136−.135|
 %𝐷𝑖𝑓𝑓 = (.338+.337)/2 ∙ 100
 %𝐷𝑖𝑓𝑓 = (𝑊+∆𝐾𝐸)/2 ∙ 100
 %𝐷𝑖𝑓𝑓 = (.136+.135)/2 ∙ 100
.001
|𝑊−∆𝐾𝐸|
|.338−.337|
.001
 %𝐷𝑖𝑓𝑓 = .1355 ∙ 100
 %𝐷𝑖𝑓𝑓 = .3375 ∙ 100
 %𝐷𝑖𝑓𝑓 = 0.738%
 %𝐷𝑖𝑓𝑓 = 0.296%

Part 2 (% Difference)
1→2
2→3
|𝑊−∆𝐸|
 %𝐷𝑖𝑓𝑓 = (𝑊+∆𝐸)/2 ∙ 100
|.059−.065|
 %𝐷𝑖𝑓𝑓 = (.072+.08)/2 ∙ 100
 %𝐷𝑖𝑓𝑓 = (𝑊+∆𝐸)/2 ∙ 100
 %𝐷𝑖𝑓𝑓 = (.059+.065)/2 ∙ 100
.006
|𝑊−∆𝐸|
|.072−.08|
.008
 %𝐷𝑖𝑓𝑓 = .062 ∙ 100
 %𝐷𝑖𝑓𝑓 = .076 ∙ 100
 %𝐷𝑖𝑓𝑓 = 9.677%
 %𝐷𝑖𝑓𝑓 = 10.526%
3→4
1→4
|𝑊−∆𝐸|
 %𝐷𝑖𝑓𝑓 = (𝑊+∆𝐸)/2 ∙ 100
|.079−.087|
 %𝐷𝑖𝑓𝑓 = (.21+.231)/2 ∙ 100
 %𝐷𝑖𝑓𝑓 = (𝑊+∆𝐸)/2 ∙ 100
 %𝐷𝑖𝑓𝑓 = (.079+.087)/2 ∙ 100
.008
|𝑊−∆𝐸|
|.21−.231|
.021
 %𝐷𝑖𝑓𝑓 = .083 ∙ 100
 %𝐷𝑖𝑓𝑓 = .2205 ∙ 100
 %𝐷𝑖𝑓𝑓 = 9.638%
 %𝐷𝑖𝑓𝑓 = 9.524%
Discussion
For the conditions used in part 1 the work energy theorem states that the work done on an object
is equal to the change in that object’s kinetic energy. There was a percentage difference less than
2.5 for each position change that was analyzed in part 1; further proving that the work energy
theorem is a valid theory. In part 2 work is equal to the sum of the change in kinetic energy and
the change in potential energy. This application of the work energy theorem was shown to be
true since there was no percentage difference greater than 11. It can be said that the percentage
differences in part 2 is expected to be greater than the ones in part 1. Since when the track is
inclined there are more forces action on the object, increasing the likelihood of errors; because
all these forces will not be considered in the calculations. The results from the experiment were
both accurate and precise as both parts had low percentage differences meaning that there were
only minute errors that could be attested to some form of friction. Also, the percentage
differences for each part of the experiment were in the same range of values, proving the values
were precise.
The equation of work establishes that if work is done on an object it means that a net force acting
on the object moved it over a distance. From newtons second law it is known that the said net
force causes the object to move with an acceleration that is equal to the force divided by the
mass. Since the object is accelerating its velocity will change over time as shown by the equations
1
of motion. Since motion energy (kinetic energy) can be defined as 2 𝑚𝑣 2 and there in a change
in velocity, the energy will also change. Therefore, the total work done on an object changes the
energy of said object.
Conclusion
The results from both parts of the experiment were consistent with the work energy theorem,
proving that the theorem works for both flat and inclined systems. The experiment also showed
in detail how and why the work done on an object affects and changes its energy.