edited PHYS Lab#6

Lab #6: Conservation of Energy PHYS-UA11

Objective and Description In this experiment, students further develop their skills with the Capstone interface in order to observe the Conservation of Energy. This lab consists of three parts. In the first part, a rubber tube is dropped through the photogate sensor that students were introduced to in previous experiments. In the second part, a 0.1kg hanging mass and the photogate are used, which students have been previously introduced to. However, they are used in a new way, as the mass is set up in such a way that it can be treated as a pendulum. Finally, in the last part, a 0.5kg weight is attached to a spring and allowed to oscillate, while its motion is detected with the motion sensor. Through each experiment, the application of the conservation of energy in a physical sense can be evaluated. Theory In previous experiments, the work-energy theorem, which shows that the integration of Newton’s Second Law yields W=∆KE, was observed. This experiment expands upon that observation to state that this work, or line integral, may have two different possibilities. In one instance, the work depends on the path and is considered non-conservative , while in the other, it is path-independent and therefore, conservative . From the second case, the work can be derived such that the potential energy is equal to the kinetic energy, or, U f + KE f = U i + KE i Where each side is the total mechanical energy, E . If E is conserved such that E f =E i , this is referred to as the conservation of energy . Procedure The procedure was followed fairly closely to that of the write-up with a few differences. For section 3, we only used heights of 15cm and 25cm and did not include 35cm. We also only used the rubber tube and not the paper tube. One significant mistake that occurred during this part happened due to

2 0.97 1.47 3 0.98 1.48 4 1.05 1.43 5 1.04 1.47 Table 3. Theoretical velocity, v f h = 7 cm h = 15 cm = = 1.17 m/s 𝑣 = 2𝑔ℎ 2(9. 8𝑚/𝑠 2 )(0. 07𝑚) = = 1.71 m/s 𝑣 = 2𝑔ℎ 2(9. 8𝑚/𝑠 2 )(0. 15𝑚) Table 4. Section 4 Figure 1. Highest point (H) 0.766m Distance= H–L =0.766m–0.367m =0.399m Lowest Point (L) 0.367m Max down v after H -0.83 m/s Amplitude=A Distance=2A A= =0.200m 0.399𝑚 2 Max v after L 0.83 m/s Table 5.

Height spring without mass: 122.35cm Height spring with mass: 65.15cm y 0 = –0.572m k= 𝑚𝑔 𝑦 0 = = –8.57 N/m (0.5𝑘𝑔)(9.8𝑚/𝑠 2 ) −0.572𝑚 y M =center of mass of 0.5kg weight= 0.024m Mass at highest point and KE=0, Energy=E H E H = 1 2 𝑘(𝑦 ? + 𝐴) 2 + 𝑚𝑔(ℎ 0 + 𝐴) = 1 2 (− 8. 57?/𝑚)(0. 024𝑚 + 0. 200𝑚) 2 + 0. 5•9. 8𝑚/𝑠 2 (0 + 0. 2 =0.765 J Mass at maximum down v after H, Energy=E 01 E 01 = 1 2 𝑚𝑣 2 + 1 2 𝑘𝑦 ? 2 + 𝑚𝑔ℎ 0 = (0.5kg)(-0.83m/s) 2 + (-8.57N/m)(.024m) 2 +(0.5kg)(9.8m/s 2 )(0) 1 2 1 2 =0.175 J Mass at lowest point and KE=0, Energy=E L E L = 1 2 𝑘(𝑦 ? − 𝐴) 2 + 𝑚𝑔(ℎ 0 − 𝐴) = 1 2 (− 8. 57?/𝑚)(0. 024𝑚 − 0. 200𝑚) 2 + 0. 5•9. 8𝑚/𝑠 2 (0 − 0 = -1.11 J Mass has maximum up v after L, Energy=E 02 E 02 = 1 2 𝑚𝑣 2 + 1 2 𝑘𝑦 ? 2 + 𝑚𝑔ℎ 0 = (0.5kg)(0.83m/s)+ (-8.57N/m)(.024m) 2 +(0.5kg)(9.8m/s 2 )(0) 1 2 1 2 =0.175 J Table 6. Questions 1. Energy is not conserved if friction or other non-conservative forces are present. Why? When a non-conservative force acts on a system, this work adds or removes mechanical energy to the system that is not accounted for in the conservation of energy. 2. Horizontal motion does not contribute to the PE. Why? Since we are only concerned with gravitational potential energy, only this y-axis is considered. This becomes more transparent through a free body diagram in which gravity is directed downward and opposing forces, be it the Normal force or frictional force if it is non-conservative, is directed upwards.

9. Does your intuition about the motion correspond to what the graphs are displaying? Yes, my intuition about the natural sinusoidal oscillations of a spring’s motion matches that of the general shape of my graphs in Figure 1. 10. Is the acceleration maximum or minimum when the velocity is zero? When the velocity reaches zero after its maximum point, the acceleration is at a minimum. On the other hand, when the velocity returns to zero after dropping to it’s minimum, the acceleration is at a maximum. In other words, the two graphs appear to be 90º out of phase. 11. When the velocity is maximum, is the acceleration maximum or minimum? When the velocity is maximum, the acceleration is zero. This is expected since acceleration is the derivative of velocity. 12. Is there kinetic energy that is not given by ½ mv 2 ? If energy is conserved, there would be no other kinetic energy. However, if friction is present, KE is being lost. 13. Why would energy vary at each point? Energy varies at each point because when the object is at its highest and at rest, kinetic energy is zero and all the energy comes from potential energy. As the object falls and h decreases, the potential energy becomes kinetic energy. Since friction is present, however, some of this kinetic energy is converted to thermal energy that is not accounted for. As such, the energy does not appear to be conserved. 14. Is energy conserved? Based on the data calculated in table 6, while the energy from the maximum up and down velocities is conserved, the energy from the maximum to minimum heights, H and L, are not conserved. 15. What factors might introduce error into your measurements and calculations?

In addition to air friction, since a photogate is being used, not using an index card or slight variations in the movement of the string my lead to errors. 16. Will friction affect your experimental results? Yes, friction will affect the experiment results. As previously explained, friction leads the the appearance of energy not being conserved. 17. Is there any evidence of friction in the curve of position vs time graph? Explain. In the position vs. time graph, the minimum distance the mass travels downwards decrease slightly with each successive cycle. This indicates that perhaps friction is present, leading to the mass not travelling as far. Error Analysis Section 3 Standard Deviations h =15 cm SD of individual trials Mean: 1.48 SD: 6.62•10 -2 SD between v t and v f Mean: 1.52 SD: 1.12•10 -1 h= 25 cm SD of individual trials Mean: 2.08 SD: 2.16•10 -1 SD between v t and v f Mean: 2.11 SD: 2.00•10 -1 Section 4 Standard Deviations h =7 cm SD of individual trials Mean: 1.01 SD: 3.56•10 -2 SD between v t and v f Mean: 1.04 SD: 7.34•10 -2