WORK AND ENERGY

1 WORK AND ENERGY Felecia Hassan, Lab Partner: Ariela Rodriguez Section:007, 10/16/2023

2 I. Objective The purpose of this lab was to examine the work- energy theorem. This theorem states that all forces that acts on a system must be equal to the difference in the total difference in kinetic energy. II. Description For the first portion of the lab a glider was placed onto an air track. A string was attached to one end of the glider and ran through the photogate with a smart pulley. The other end of the string was passed through the photogate with the smart pulley and 3 rubber bands. The last rubber band was attached to a force sensor, that was clamped near to the floor. The rubber bands were stretched (by moving the glider backward ) just before it blocked the photogate, and then let go. During these trials the change in kinetic energy was measured. The velocity and force was also measured my Capstone software by displaying a curve that measured the relationship between force and work. In the following portion of the lab , we examined the relationship between tension and a system of two masses. The glider remained on the air track and a string was attached to the end of the glider. The other end of the string was ran through the photogate and attached to a mass. The rubber bands were stretched (by moving the glider backward ) just before it blocked the photogate, and then let go. Upon release, the velocity of each trial was determined by looking at the relationship between distance and time.

4 Trial 1 : ∆ KE = 1 2 m ( V f 2 ) − 1 2 m ( V i 2 ) ∆ KE = 1 2 ( 0.4468 k g ) ( 0.50136 2 ) − 1 2 0.446.8 kg ( 0.12666 2 ) ∆ KE = 0 . 05273 Error : | W − ∆ KE | Error : | 0.06082 − 0.05273 | Error : 0.008609 Trial 2 : ∆ KE = 1 2 m ( V f 2 ) − 1 2 m ( V i 2 ) ∆ KE = 1 2 ( 0.4468 k g ) ( 0.50784 2 ) − 1 2 0.4468 k g ( 0.08310 2 ) ∆ KE = 0 . 05607 Error : | W − ∆ KE | Error : | 0.06121 − 0 . 05607 | Error : 0.00514 Trial 3 : ∆ KE = 1 2 m ( V f 2 ) − 1 2 m ( V i 2 ) ∆ KE = 1 2 ( 0.4468 g ) ( 0.60032 2 ) − 1 2 0.4468 kg ( 0.06550 2 ) ∆ KE = ¿ 0.09584 Error : | W − ∆ KE | Error : | 0.09141 − 0.09584 | Error : 0.00443 Trial 4 : ∆ KE = 1 2 m ( V f 2 ) − 1 2 m ( V i 2 )

5 ∆ KE = 1 2 ( 0.4468 kg ) ( 0.59515 2 ) − 1 2 0.4468 kg ( 0.09540 2 ) ∆ KE = ¿ 0.07709 Error : | W − ∆ KE | Error : | 0.07752 − 0.07709 | Error : 0.000443 Trial 5 : ∆ KE = 1 2 m ( V f 2 ) − 1 2 m ( V i 2 ) ∆ KE = 1 2 ( 0.4468 kg ) ( 0.59515 2 ) − 1 2 0.4468 kg ( 0.09540 2 ) ∆ KE = ¿ 0.07713 Error : | W − ∆ KE | Error : | 0.07366 − 0.07713 | Error : 0.00347 PART 2 Trial Initial Position (m) Final Position (m) Initial Velocity (m/s) Final Velocity (m/s) 1 0.015 0.555 0.0826 1.01218 2 0.015 0.675 0.15064 1.12907 3 0.015 0.585 0.09464 1.0098 4 0.015 0.525 0.14232 0.99467 5 0.015 0.645 0.24512 1.12136 a. PART 2 ∆ KE Trial 1 : ∆ KE = 1 2 m ( V f 2 ) − 1 2 m ( V i 2 ) ∆ KE = 1 2 ( 0.4468 kg ) ( 1.01218 2 ) − 1 2 0.446.8 kg ( 0.0826 2 ) ∆ KE = 0 . 22735

7 Error : | W − ∆ KE | Error : | 2.793 − 0.22579 | Error : 2.5672 Trial 4 : ∆ KE = 1 2 m ( V f 2 ) − 1 2 m ( V i 2 ) ∆ KE = 1 2 ( 0.4468 kg ) ( 0.99467 2 ) − 1 2 0.4468 kg ( 0.14232 2 ) ∆ KE = 0.00458 Work ( w ) = ¿ M1g(x((final)- x( initial) ) ( w ) = ¿ 0.50 kg (9.81 m/s^2)( 0.525m- 0.015m) ( w ) = 2.499 Error : | W − ∆ KE | Error : | 2.499 − 0.00458 | Error : 2.4944 Trial 5 : ∆ KE = 1 2 m ( V f 2 ) − 1 2 m ( V i 2 ) ∆ KE = 1 2 ( 0.4468 kg ) ( 1.12136 2 ) − 1 2 0.4468 kg ( 0.24512 2 ) ∆ KE = ¿ 0.3155 Work ( w ) = ¿ M1g(x((final)- x( initial) ) ( w ) = ¿ 0.50 kg (9.81 m/s^2)( 0.645 m- 0.015m) ( w ) = 3.087 Error : | W − ∆ KE | Error : | 3.087 − 0.3155 | Error : 2.7715

8 VI. Error Analysis PART 1 ERROR ANAYLIS STANDARD DEVIATION μ ( mean of ∆ KE )= 0.071772 ( i ) − μ X ¿ ¿ ¿ 2 Σ ¿ ¿ σ = √ ¿ σ = √ 6.38 ( 10 ) 4 5 σ = 0.1129 μ ( WORK ) = 0.072924 Error : | μW − μ ∆KE | Error : | 0.072924 − 0.071772 | Error : 0.001152 PART 2 ERROR ANAYLIS STANDARD DEVIATION μ ( mean of ∆ KE )= 0.071772 ( i ) − μ X ¿ ¿ ¿ 2 Σ ¿ ¿ σ = √ ¿ σ = √ 6.38 ( 10 ) 4 5 σ = 0.1129 μ ( WORK ) = 2.853 Error : | μW − μ ∆KE | Error : | 2.853 − 0.071772 | Error : 2.7812 VII. Analysis ( Error Analysis continued )

10 a. External forces on M1 include gravitational force, and air resistance that opposes the motion of M1. 5. Compare the work done with the change in KE for these three items. Is there a difference in value? Explain. a. As mentioned before there will be a difference in these values because work includes several types of energy including kinetic energy. 6. The work done on M1 is (M1g − T)(xf − xi). Explain the minus sign before the T. The work done on M2 is T(xf − xi). a. T is negative because the motion (direction) of M1 is falling downwards. Inversely, the tension of the string is applied upwards. This means that T (tension) opposes the direction of the object 7. The work done on the system is M1g(xf − xi). Explain clearly why T does not appear in this expression a. This is because tension is an internal force in this context since is acts on both M1 and M2. Since this tensions is applied to both masses, the tension applied to M1 is equal in magnitude and has an opposing sign when compared to the tension applied to M2. So, these tensions would cancel each other out. IX. Conclusion We were able to successfully compare the change in kinetic energy to the area under the position v. force curve, and found that there was very little variance between the calculated change in kinetic energy ( the calculated value of work) and the measured work done by the system ( area under the position v. force curve). However, when considering our results from the second portion of the lab we observed a much larger discrepancy between the calculated change

11 in kinetic energy and the measured work done by the system. This is likely due to the factors mentioned in error analysis in the section above. Nonetheless partner and I were able to observe that the change in kinetic energy over a particular displacement can be compared to the net force applied to that object over that particular displacement.