Please I need help with my Physics Lab analysis question. I need a full explanti
ID: 1495125 • Letter: P
Question
Please I need help with my Physics Lab analysis question. I need a full explantion of every answer because this is a lab report.
Subject is Conservation of Mechanical Energy.
Free body diagram of the cart which is traveling in the straight track:-
Q.1:Using your diagram from question 1, sum the forces acting perpendicular to the track. Sum the forces acting parallel to the track. Using these equations, find the formula for the acceleration of the cart down the track. Use this formula to calculate the acceleration in this table. (Cart weight = 0.5610kg)
t(s)
x(m)
v(m/s)
height(m)
PE(J)
KE(J)
TME(J)
2.5666
0.4929
0.3463
0.0143
0.0784
0.0336
0.1121
2.6000
0.5037
0.3437
0.0140
0.0769
0.0331
0.1100
2.6333
0.5160
0.3349
0.0137
0.0751
0.0315
0.1065
2.6666
0.5262
0.3221
0.0134
0.0736
0.0291
0.1027
2.7000
0.5370
0.3247
0.0131
0.0721
0.0296
0.1016
2.7333
0.5478
0.3309
0.0128
0.0705
0.0307
0.1012
2.7666
0.5589
0.3410
0.0125
0.0689
0.0326
0.1015
2.8000
0.5702
0.3563
0.0122
0.0673
0.0356
0.1029
2.8333
0.5833
0.3560
0.0119
0.0654
0.0356
0.1009
2.8666
0.5942
0.3474
0.0116
0.0638
0.0339
0.0976
2.9000
0.6059
0.3510
0.0113
0.0621
0.0346
0.0967
2.9333
0.6179
0.3490
0.0110
0.0604
0.0342
0.0945
2.9666
0.6291
0.3514
0.0107
0.0588
0.0346
0.0934
3.0000
0.6411
0.3607
0.0104
0.0570
0.0365
0.0935
3.0333
0.6532
0.3686
0.0101
0.0553
0.0381
0.0934
3.0666
0.6656
0.3763
0.0097
0.0535
0.0397
0.0932
3.1000
0.6781
0.3873
0.0094
0.0517
0.0421
0.0938
3.1333
0.6923
0.3767
0.0090
0.0496
0.0398
0.0894
3.1666
0.7030
0.3732
0.0087
0.0481
0.0391
0.0871
3.2000
0.7158
0.4033
0.0084
0.0462
0.0456
0.0919
3.2333
0.7308
0.4117
0.0080
0.0441
0.0476
0.0916
3.2666
0.7440
0.4026
0.0077
0.0422
0.0455
0.0876
3.3000
0.7572
0.4013
0.0073
0.0403
0.0452
0.0854
3.3333
0.7707
0.4005
0.0070
0.0383
0.0450
0.0833
3.3666
0.7838
0.4042
0.0066
0.0364
0.0458
0.0822
3.4000
0.7976
0.4095
0.0063
0.0344
0.0470
0.0814
3.4333
0.8110
0.4157
0.0059
0.0325
0.0485
0.0810
3.4666
0.8251
0.4288
0.0055
0.0304
0.0516
0.0820
3.5000
0.8400
0.4312
0.0051
0.0283
0.0522
0.0804
3.5333
0.8539
0.4315
0.0048
0.0263
0.0522
0.0785
3.5666
0.8686
0.4339
0.0044
0.0242
0.0528
0.0770
3.6000
0.8829
0.4352
0.0040
0.0221
0.0531
0.0752
3.6333
0.8976
0.4382
0.0036
0.0200
0.0539
0.0738
3.6666
0.9120
0.4432
0.0033
0.0179
0.0551
0.0730
3.7000
0.9270
0.4538
0.0029
0.0157
0.0578
0.0735
3.7333
0.9424
0.4618
0.0025
0.0135
0.0598
0.0733
3.7666
0.9580
0.4629
0.0020
0.0112
0.0601
0.0714
3.8000
0.9731
0.4672
0.0016
0.0091
0.0612
0.0703
3.8333
0.9892
0.4693
0.0012
0.0067
0.0618
0.0685
3.8666
1.0045
0.4701
0.0008
0.0045
0.0620
0.0665
3.9000
1.0204
0.4745
0.0004
0.0022
0.0632
0.0654
3.9333
1.0359
0.4851
0.0000
0.0000
0.0660
0.0660
t(s)
x(m)
v(m/s)
height(m)
PE(J)
KE(J)
TME(J)
2.5666
0.4929
0.3463
0.0143
0.0784
0.0336
0.1121
2.6000
0.5037
0.3437
0.0140
0.0769
0.0331
0.1100
2.6333
0.5160
0.3349
0.0137
0.0751
0.0315
0.1065
2.6666
0.5262
0.3221
0.0134
0.0736
0.0291
0.1027
2.7000
0.5370
0.3247
0.0131
0.0721
0.0296
0.1016
2.7333
0.5478
0.3309
0.0128
0.0705
0.0307
0.1012
2.7666
0.5589
0.3410
0.0125
0.0689
0.0326
0.1015
2.8000
0.5702
0.3563
0.0122
0.0673
0.0356
0.1029
2.8333
0.5833
0.3560
0.0119
0.0654
0.0356
0.1009
2.8666
0.5942
0.3474
0.0116
0.0638
0.0339
0.0976
2.9000
0.6059
0.3510
0.0113
0.0621
0.0346
0.0967
2.9333
0.6179
0.3490
0.0110
0.0604
0.0342
0.0945
2.9666
0.6291
0.3514
0.0107
0.0588
0.0346
0.0934
3.0000
0.6411
0.3607
0.0104
0.0570
0.0365
0.0935
3.0333
0.6532
0.3686
0.0101
0.0553
0.0381
0.0934
3.0666
0.6656
0.3763
0.0097
0.0535
0.0397
0.0932
3.1000
0.6781
0.3873
0.0094
0.0517
0.0421
0.0938
3.1333
0.6923
0.3767
0.0090
0.0496
0.0398
0.0894
3.1666
0.7030
0.3732
0.0087
0.0481
0.0391
0.0871
3.2000
0.7158
0.4033
0.0084
0.0462
0.0456
0.0919
3.2333
0.7308
0.4117
0.0080
0.0441
0.0476
0.0916
3.2666
0.7440
0.4026
0.0077
0.0422
0.0455
0.0876
3.3000
0.7572
0.4013
0.0073
0.0403
0.0452
0.0854
3.3333
0.7707
0.4005
0.0070
0.0383
0.0450
0.0833
3.3666
0.7838
0.4042
0.0066
0.0364
0.0458
0.0822
3.4000
0.7976
0.4095
0.0063
0.0344
0.0470
0.0814
3.4333
0.8110
0.4157
0.0059
0.0325
0.0485
0.0810
3.4666
0.8251
0.4288
0.0055
0.0304
0.0516
0.0820
3.5000
0.8400
0.4312
0.0051
0.0283
0.0522
0.0804
3.5333
0.8539
0.4315
0.0048
0.0263
0.0522
0.0785
3.5666
0.8686
0.4339
0.0044
0.0242
0.0528
0.0770
3.6000
0.8829
0.4352
0.0040
0.0221
0.0531
0.0752
3.6333
0.8976
0.4382
0.0036
0.0200
0.0539
0.0738
3.6666
0.9120
0.4432
0.0033
0.0179
0.0551
0.0730
3.7000
0.9270
0.4538
0.0029
0.0157
0.0578
0.0735
3.7333
0.9424
0.4618
0.0025
0.0135
0.0598
0.0733
3.7666
0.9580
0.4629
0.0020
0.0112
0.0601
0.0714
3.8000
0.9731
0.4672
0.0016
0.0091
0.0612
0.0703
3.8333
0.9892
0.4693
0.0012
0.0067
0.0618
0.0685
3.8666
1.0045
0.4701
0.0008
0.0045
0.0620
0.0665
3.9000
1.0204
0.4745
0.0004
0.0022
0.0632
0.0654
3.9333
1.0359
0.4851
0.0000
0.0000
0.0660
0.0660
Normal force (Normal - Weight) Normal force (Normal-weight) m g is also Ymass-gravity be the weight of an object massxgravity considered to m 9 Friction force Friction =m×N m-g-cos Fi- meg-cos .Explanation / Answer
From what I see the numerical answer is already on the table so I will to justify the results.
In the table we are able to see different aspects in the analysis of mechanical energy, we are able to notice that while the potential energy is reducing the kinetic energy grows consecuence of the acceleration of caused by the gravity, but we are also able to notice that although the kinetic energy grows while the potencial is reducing we can see that the total energy of the system doesn't remain constant, this means that the system have losses an it makes sense considering that there's friction in the system, from all this data we are able to notice that the effect that the gravity have on the mass is greater than the effect of the friction.
For the theorical value of the acceleration the value of the friction coefficient is needed, but with the table we should be able to define average experimental values:
The acceleration (Vf-Vi)/(tf-ti)=0.5043 m/s^2
The average value of the experimental friction could be deterined by the energy loss knowing that it would fr(Xf-Xi)=Ei-Ef => fr=(Ef-Ei)/(Xf-Xi)=3.3738*10^-2
Knowing fr we can define Uk fr