Conservation of Energy

pdf

School

San Antonio College *

*We aren’t endorsed by this school

Course

1611

Subject

Mechanical Engineering

Date

Dec 6, 2023

Type

pdf

Pages

Uploaded by CorporalBook6857

1 Analysis of Conservation of Energy Lab Name Course/Section PHY1611 Table 1 ! ! = 1.00 meter Setting 1 Setting 2 Setting 3 +, " 1.320 m 1.722 m 2.152 m +, # 1.318 m 1.726 m 2.172 m +, $ 1.329 m 1.734 m 2.173 m +, %&' 1.322 m 1.727 m 2.166 m - 0.4515 sec 0.4515 sec 0.4515 sec . ( 2.93 m/s 3.83 m/s 4.80 m/s 1. Calculate the average x-displacement for each setting, and enter your answers in Table 1 . - 132.0+131.8+132.9/3= 132.2 cm converted to meters is 1.322 m - 172.2+172.6+173.4/3= 172.7 cm converted to meters is 1.727 m - 215.2+217.2+217.3/3= 216.6 cm converted to meters is 2.166 m I converted everything to meters after doing #3 of this section due to it being easier to calculate m/s instead of cm/s. Measurements originally taken in cm. 2. Using the equation ! = ! ! + 0 !) 1 − * + 31 + calculate the time of flight for each setting, and enter your answers in Table 1 . (6 points) - ½ (9.8m/s^2)t^2=1.00 t = 0.4515s 3. Using the equation 0 ! = ,- !"# . calculate the initial velocity for each setting and enter your answers in Table 1 . (6 points) - 1.322/0.4515 = 2.93 m/s - 1.727/0.4515 = 3.83 m/s - 2.166/0.4515 = 4.80 m/s

2 Table 2: Conservation of Mechanical Energy ! ! = 1.00 meter # = 16.7( Setting 1 Setting 2 Setting 3 4 /" 0.47 m 0.67 m 1.41 m 4 /# 0.48 m 0.68 m 1.46 m 4 /$ 0.47 m 0.68 m 1.44 m 4 /0 0.48 m 0.64 m 1.43 m 4 /1 0.47 m 0.66 m 1.49 m 4 /2%&' 0.474 m 0.666 m 1.446 m 56 / 0.0776 J 0.1090 J 0.2367 J . ( 2.93 m/s 3.83 m/s 4.80 m/s 76 ( 0.0717 J 0.1225 J 0.1924 J 1. Calculate the average final height for each setting, and the Potential Energy the projectile has at its maximum height. Then record those answers in Table 2 . (20 points) PEf=mgYf ((Mass= 16.7 g; converted to kg is 0.0167kg Gravity 9.8m/s^2)) - 0.0167(9.8)(0.474)= 0.0776 J - 0.0167(9.8)(0.666)= 0.1090 J - 0.0167(9.8)(1.446)= 0.2367 J 2. Transfer the values for the velocities for each setting from Table 1 to Table 2 , then calculate the initial Kinetic Energies for each setting. Record those answers in Table 2 . (20 points) KEi= ½ mVi^2 - ½ ( 0.0167 )( 2.93 )^2 = 0.0717 J - ½ ( 0.0167 )( 3.83 )^2 = 0.1225 J - ½ ( 0.0167 )( 4.80 )^2 = 0.1924 J 3. Calculate the % difference between the initial Kinetic Energy and the final Potential Energy for each setting. (8 points) - (0.0776-0.0717)/((0.0776+0.0717)/2)(100%)= 7.91% - (0.1090-0.1225)/((0.1090+0.1225)/2)(100%)= 11.66% - (0.2367-0.1924)/((0.2367+0.1924)/2)(100%)= 20.65% 4. Was the mechanical energy conserved for each setting? (8 points) - It became evident that the starting mechanical energy and the final mechanical energy were not identical for any of the parameters. This implies that the system does not preserve mechanical energy.

3 5. Give 3 reasons why we have obtained percent difference (do not include rounding/calculation errors, bad equipment, etc). (6 points) - Differences between initial and end energy might result from system effects from outside influences like friction or air resistance. - Energy can change or disappear into sound or heat, leading to a discrepancy between what is expected and what is seen. - There can be % disparities in trials because simplified theoretical models don't always consider all the intricacies of the real world. 6. Do the results of our experiment support the theory? Explain your answer. (6 points) - Yes, the results of our experiment do support the theory because for the most part, the mechanical energy was conserved. “conservation of energy.” Now I know our values were not exact but they were fairly close or within reason when you take into account the different factors that could throw our data off.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version