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SENIOR HIGH SCHOOL REPORTThis was PART B of a Physics Lab (part A should be below)- Conservation of Energy- Elastic Potential Energy. This was just your typical conservation of energy lab where the energy is not conserved due to a non ideal spring, with the slight exception that the wanted you to overcome the 'non ideality' LOL of the spring in order to come up with the correct answer. Probably not that hard when I think about it now but I remember it took me longer than I had expected.If u want the original just msg me where u want it sent-
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Mark Riley 3107631608
RESULTS/CALULATIONS PAGE3
TABLE1a.
Maximum extension of spring when a 1.00kg weight was added -
Extension of spring when a 9.81N force is applied Force applied Fw = (1.00x9.81) = 9.81N
1st Trial 2nd Trial 3rd Trial 4th Trial 5th Trial 6th Trial 7th Trial 8th Trial 9th Trial 10th Trial
0.505m 0.505m 0.500m 0.500m 0.510m 0.508m 0.510m 0.512m 0.510m 0.510m
TABLE2a.
Extension of spring (x) versus Mass (m) and Force Applied (Fw)
Extension (m) 0.000m 0.000m 0.010m 0.046m 0.090m 0.111m 0.145m 0.181m 0.216m 0.251m
Mass (kg) 0.100kg 0.200kg 0.300kg 0.400kg 0.500kg 0.600kg 0.700kg 0.800kg 0.900kg 1.00kg
Force Applied 0.981N 1.96N 2.94N 3.92N 4.91N 5.88N 6.86N 7.84N 8.83N 9.81N
LIMITATIONS OF THE EQUIPMENT MUST BE TAKEN INTO CONSIDERATION FOR ALL MEASUREMENTS OF
EXTENSION AND MASS IN TABLE 1A. AND 2A.
PRECISION OF RULER 1mm ∴ ±0.5mm PRECISION OF MASS 1gram ∴ ±0.5g
GRAPH A.
y = 31.98x + 2.0376
0.000
2.000
4.000
6.000
8.000
10.000
12.000
0.000 0.050 0.100 0.150 0.200 0.250 0.300
Forc
e (N
)
Extension (m)
Force versus Extension
Mark Riley 3107631608
RESULTS/CALULATIONS PAGE4
GRAPH B.
Blue values represent data that was disregarded when plotting the trend line
TABLE3b.
Extension (m x 10-2) Force (N)
0.00 0.981
0.00 1.96
1.00 2.94
4.60 3.92
9.00 4.91
11.1 5.89
14.5 6.87
18.1 7.85
21.6 8.83
25.1 9.81
y = 28.901x + 2.5832
0.000
2.000
4.000
6.000
8.000
10.000
12.000
0.000 0.050 0.100 0.150 0.200 0.250 0.300
Forc
e (N
)
Extension (m )
Force versus Extension
Mark Riley 3107631608
RESULTS/CALULATIONS PAGE5
CALC1a. The average maximum extension (x) of 1kg weight using values from TABLE1a.
𝟓𝟎.𝟓𝒄𝒎 + 𝟓𝟎.𝟓𝒄𝒎 + 𝟓𝟎.𝟎𝒄𝒎 + 𝟓𝟎.𝟎𝒄𝒎 + 𝟓𝟏.𝟎𝒄𝒎 + 𝟓𝟎. 𝟖𝒄𝒎 + 𝟓𝟏.𝟎𝒄𝒎 + 𝟓𝟏.𝟐𝒄𝒎 + 𝟓𝟏.𝟎𝒄𝒎 + 𝟓𝟏.𝟎𝒄𝒎
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒕𝒓𝒊𝒂𝒍𝒔 𝟏𝟎.𝟎
𝟓𝟎𝟕𝒄𝒎
𝟏𝟎.𝟎= 𝟓𝟎.𝟕𝒄𝒎 = 𝟎. 𝟓𝟎𝟕𝒎𝒆𝒕𝒓𝒆𝒔 𝒐𝒓 𝟓.𝟎𝟕𝐱𝟏𝟎−𝟏𝒎𝒆𝒕𝒓𝒆𝒔
CALC1b. Uncertainty in the measure of extension(x) for the 1kg weight
50.7 − 50.5 = 𝟎. 𝟐 50.7 − 50.5 = 𝟎.𝟐 50.7 − 50 = 𝟎. 𝟕 50.7 − 50 = 𝟎. 𝟕
51.0 − 50.7 = 𝟎. 𝟑 50.8 − 50.7 = 𝟎.𝟏 51.0 − 50.7 = 𝟎. 𝟑 51.2 − 50.7 = 𝟎.𝟓
51.0 − 50.7 = 𝟎. 𝟑 51 − 50.7 = 𝟎. 𝟑
𝟎.𝟐𝒄𝒎 + 𝟎.𝟐𝒄𝒎 + 𝟎.𝟕𝒄𝒎 + 𝟎.𝟕𝒄𝒎 + 𝟎. 𝟑𝒄𝒎 + 𝟎. 𝟏𝒄𝒎 + 𝟎. 𝟑𝒄𝒎 + 𝟎. 𝟓𝒄𝒎 + 𝟎.𝟑𝒄𝒎 + 𝟎.𝟑𝒄𝒎
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒕𝒓𝒊𝒂𝒍𝒔 𝟏𝟎
𝟑.𝟔𝒄𝒎
𝟏𝟎= 𝟎. 𝟑𝟔𝒄𝒎 = 𝟎.𝟎𝟎𝟑𝟔𝒎𝒆𝒕𝒓𝒆𝒔 𝒐𝒓 𝟑.𝟔𝐱𝟏𝟎−𝟑𝒎𝒆𝒕𝒓𝒆𝒔
∆x = ± 3.6mm
CALC1c. Express this uncertainty as a percentage
𝟎. 𝟑𝟔𝒄𝒎
𝟓𝟎.𝟕𝒄𝒎 𝐱 𝟏𝟎𝟎 = 𝟎. 𝟕𝟏%
Mark Riley 3107631608
RESULTS/CALULATIONS PAGE6
CALC3a. Spring constant = the gradient of a Force applied vs extension graph. Using the two
points that best fit the trend line in respect to GRAPH B.
𝐘𝟏− 𝐘𝟐
𝐗𝟏− 𝐗𝟐 ∴
𝟖.𝟖𝟑 – 𝟑.𝟗𝟐
𝟎.𝟐𝟏𝟔−𝟎.𝟎𝟒 𝟔= 𝟐𝟖.𝟗 𝐊 = 𝟐𝟖.𝟗 𝐉
CALC4a. Loss of Gravitation Potential Energy (GPE) using height (x) value from CALC1a. All
GPE is transferred at the relative height of zero.
𝑮𝑷𝑬 = 𝒎𝒈𝒉 𝒎 = 𝒎𝒂𝒔𝒔 𝒈 = 𝒈𝒓𝒂𝒗𝒊𝒕𝒚 (𝟗.𝟖𝟏) h=height (extension in this case)
𝑮𝑷𝑬 𝒕𝒓𝒂𝒏𝒔𝒇𝒆𝒓𝒓𝒆𝒅 = 𝟏.𝟎𝟎𝒌𝒈 𝐱 𝟗.𝟖𝟏 𝐱 𝟎.𝟓𝟎𝟕𝒎 = 𝟒.𝟗𝟕 𝑱
CALC4b. Gain in Elastic Potential Energy (EPE) using K value from CALC3a. and the X value
from CALC1a.
𝑬𝑷𝑬 = 𝟏
𝟐𝒌𝒙𝟐 𝒌 = 𝒔𝒑𝒓𝒊𝒏𝒈 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒙 = 𝒆𝒙𝒕𝒆𝒏𝒔𝒊𝒐𝒏
𝑬𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 =𝟏
𝟐 𝐱 𝟐𝟖.𝟗 𝐱 𝟎.𝟓𝟎𝟕𝟐 = 𝟑.𝟕𝟏 𝑱
CALC5a. Difference between loss of GPE and gain of EPE using values from CALC4a. &
CALC4b.
𝑮𝑷𝑬− 𝑬𝑷𝑬 = 𝑫𝑰𝑭𝑭𝑬𝑹𝑬𝑵𝑪𝑬 𝟒.𝟗𝟕𝑱− 𝟑.𝟕𝟏𝑱 = 𝟏.𝟐𝟔𝑱
Mark Riley 3107631608
RESULTS/CALCULATIONS PAGE7
CALC5b. Difference between the two energies as a percentage in relation to GPE
𝟏. 𝟐𝟔
𝟒. 𝟗𝟕 𝐱 𝟏𝟎𝟎 = 𝟐𝟓.𝟒%
REF6a. Comparison between CALC5b. and CALC1c.
𝟐𝟓.𝟒% 𝒗𝒆𝒓𝒔𝒖𝒔 𝟎.𝟕𝟏%
CALC7a. Area underneath a Force vs Extension graph should equal the work done.
Work done in respect to GRAPH B
2.58 x 0.251 = 𝟎.𝟔𝟒𝟖
9.81 − 2.58 x0.251
2= 𝟎.𝟗𝟎𝟕
𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝟎. 𝟔𝟒𝟖 + 𝟎. 𝟗𝟎𝟕 = 𝟏.𝟓𝟔𝑱
CALC7b. The actual Relationship between Force applied and Extension. Refer to the
equation to the trend line in GRAPH B. F = kx + F0
𝒀 = 𝒎𝒙 + 𝒄 𝒐𝒓 𝑭 = 𝒌𝒙 + 𝑭𝟎
𝑭 = 𝟐𝟖.𝟗𝒙 + 𝟐. 𝟓𝟖
Mark Riley 3107631608
RESULTS/CALCULATIONS PAGE8
CALC9a. The actual relationship between F and x for this spring was found in the previous
calculation CALC7b. Express this relationship in terms of area and substitute into
the expression EPE = ½kx2 . Where x will be the best estimate of the maximum
extension of the falling 1kg mass.
𝑬𝑷𝑬 = 𝟏
𝟐𝒌𝒙𝟐 + 𝑭𝒐𝒙 ∴
𝟏
𝟐𝟐𝟖.𝟗𝐱𝟎.𝟓𝟎𝟕𝟐 + 𝟐.𝟓𝟖𝐱𝟎.𝟓𝟎𝟕 = 𝟓.𝟎𝟐𝑱
CAL9b. Compare this energy to the Energy calculated for the loss of GPE CALC4a.
𝑬𝑷𝑬 𝒗𝒔 𝑮𝑷𝑬 𝟓.𝟎𝟐𝑱 𝒗𝒔 𝟒.𝟗𝟕𝑱
4.97J – 5.02J = -0.051
Difference between values = 0.051J
CALC9c. Express the difference found in the previous calculation as a percentage in respect
to GPE.
𝟎. 𝟎𝟓𝟏
𝟒. 𝟗𝟕 𝒙 𝟏𝟎𝟎 = 𝟏. 𝟎𝟑%
CALC9d. Compare this percentage difference to the percentage uncertainty in the
measurement of distance/extension value for x which is 0.71% CALC1c.
𝟏. 𝟎𝟑% 𝒗𝒔 𝟎.𝟕𝟏%
This time the percentage uncertainty for measurement of the extension can pretty
much account for the error in calculation of EPE using the equation from CALC9c.
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