PHYS 253 - Module 2 - Lab 6

Newton’s Laws EXPERIMENT 1: NEWTON’S FIRST LAW OF MOTION Data Sheet Table 1. Motion of Water Observations Motion Observations a  Water sloshed towards my hand that accelerated the container b  Water was semi-stable through the motion c  Right turn: Water sloshed left Left turn: Water sloshed right d  Water sloshed forward Table 2. Observations after Flicking Notecard Off of Cup Trial Observations 1  Washer fell into the cup 2  Washer fell into the cup 3  Washer flew off the cup with the notecard 4  Washer fell into the cup 5  Washer flew off the cup with the notecard Post-Lab Questions 1. Explain how your observations of the water and washer demonstrate Newton’s law of ©eScience Labs, 2014

Newton’s Laws 2. Compare the magnitude of the forces on both spring scales after you pulled the 10 N spring scale suspended on the 5 N spring scale. Both scales showed 5N force. 3. Use Newton’s 3rd Law to explain your observations in Questions 1 and 2. This is due to Newton’s 3rd law of motion, which states that for all applied forces, there is an equal magnitude and opposite direction force response. When one scale pulls on the other with an applied force, there is an equal and opposite force as a reaction. 4. Compare the force on the 10 N spring scale when it was directly attached to the 0.5 kg mass and when there was a string between them. The scales showed 5N for both tests. 5. Compare the force on the two spring scales in Steps 5 and 6. What can you conclude about the tension in a string?  The tension in the string is due to the applied force created by the hanging mass. The string itself does not add any extra force to the scale other than the added weight (which in this case was negligible) so it only transmits the force of the hanging 0.5kg mass via tension. The tension equals the force of the mass multiplied by the gravitational acceleration acting on the system. ©eScience Labs, 2014

Newton’s Laws φ = a r (Angular acceleration of pulley) a = g ( m 2 − m 1 ) M 2 + m 1 + m 2 (M = pulley mass) 4. Set the two resulting expressions for the force of tension from Part 1 equal to one another (as long as the string does not stretch, the magnitude of the acceleration in each equation is the same). Replace F g1 and F g2 with M 1 and M 2 , respectively. Solve the resulting equation for a . Then, go back to Question 3 and solve for the F T . Repeat this with the resulting expressions from Part 2. a = ( m 2 ×g ) −( m 1 ×g ) m 1 + m 2 5. Calculate the acceleration for the two sets of data you recorded and compare these values to those obtained by measuring distance and time using percent error. Cite two factors that may cause discrepancies between the two values. Set #1: 0.654 m s 2 Set #2: 1.962 m s 2 6. Calculate the tension in the string for the falling washers. Show all calculations. From these two values, and the one where the masses were equal, what trend do you observe in the tension in the string as the acceleration increases? Set #1: F T 1 = m 1 ( g + a ) F T 2 = m 2 ( g − a ) F T 1 = 0.073248 N F T 2 = 0.073248 N Set #2: F T 1 = m 1 ( g + a ) F T 2 = m 2 ( g − a ) F T 1 = 0.070632 N F T 2 = 0.070632 N Observation: As acceleration increases, tension in the line appears to decrease. From these results, we can conclude that tension and acceleration, for this system, are inversely proportional. ©eScience Labs, 2014