Alissa Castano- PHYS1P91-Lab5

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Brock University *

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1P91

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Physics

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Apr 3, 2024

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6

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TABLES: 50g, 100g, 150g UNIT VALUE UNCERTAINTY iOLab Weight g 200.7 0.5 UNIT VALUE UNCERTAINTY Trial 1 weight (with iOLab) N 3.14 0.03 Trial 2 Weight (with iOLab) N 3.95 0.02 Trial 3 Weight (with iOLab) N 4.13 0.03 UNIT VALUE UNCERTAINTY Pulling Force iOLab N 0.31 0.02 Pulling Force Trial 1 N 0.41 0.03 Pulling Force Trial 2 N 0.49 0.03 Pulling Force Trial 3 N 0.53 0.03 UNIT VALUE UNCERTAINTY Slope unitless 0.1245 N/A Y-intercept N -0.0204 N/A UNIT VALUE UNCERTAINTY Ramp Angle degrees 10 0.05 UNIT VALUE UNCERTAINTY A_up m/s 2 -1.109 0.006 A_down m/s 2 -0.520 0.003 μ _k N/A 0.03046 N/A UNIT VALUE UNCERTAINTY A_up m/s 2 -1.00 0.01 A_down m/s 2 -0.549 0.007
μ_ k N/A 0.02369 N/A UNIT VALUE UNCERTAINTY A_up m/s 2 -1.117 0.004 A_down m/s 2 -0.537 0.004 μ_ k N/A 0.03004 N/A UNIT VALUE UNCERTAINTY Average μ_ k N/A 0.028 0.004 FIGURES: Figure 1: Force-time graph of iOlab being pulled with no added weight.
Figure 2: Linear relationship between weight of iOlab and pulling force. Figure 3: Position-time graph of trial one, iOlab ramp experiment. Figure 4: Velocity-time graph of trial one, iOlab ramp experiment.
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Figure 5: Acceleration-time graph of trial one, iOlab ramp experiment. Figure 6: Free body diagrams of iOlab rolling up and down ramp.
DISCUSSION: What does the value of A represent? What does the Y intercept represent, what should it be? The linear equation (y= Ax + b) for the slope was found to be 0.1245, and the y-intercept -0.0204. The value of A (0.1245) represents the relationship between the pulling force necessary to drag the iOlab, and the weight of the iOlab (including the added masses). In this equation, A also represents the coefficient of friction multiplied by the force applied on the block. As for the y-intercept (b), it represents how much initial pulling force is needed to pull the iOlab with no additional weight on it. The slightly negative y-intercept (-0.0204) implies that despite there being no weight, there is still some force needed to drag the iOlab. Can you identify the point from the position graph where the maximum height was reached? Can you identify that same point from the velocity graph? Figure 3 illustrates the position-time graph for the motion of the iOlab on the ramp. As seen in the figure, there is an obvious peak in the graph representing the highest point reached by the iOlab. The highest point on the position-time graph is the highest point reached by the iOlab before it rolled down the ramp. As for the velocity-time graph, you cannot truly identify the highest point reached by the iOlab, because the velocity graph does not display the position of the iOlab, only its velocity. That being said, one can identify the time (in seconds), when the iOlab reached its highest point. This would be when a point on the velocity-time graph touches zero (see figure 4). As the iOlab reaches its peak after rolling up as high as it can, it changes direction (rolls back down). This change makes the velocity zero. Therefore, it is possible to identify the time when iOlab reached zero on the velocity graph (when y=0). When comparing figure 3, and figure 4, it's possible to see that both graphs illustrate that the iOlab reached its peak point right before 1 second on both graphs. Which coefficient of kinetic friction less between your two experiments? Why might this be? For the first experiment (dragging the iOlab on the table), the coefficient of kinetic friction was the value of A in the linear equation, so 0.1245. For the second experiment (3 trials pushing the iOlab up a ramp), the average coefficient of kinetic friction of all three trials was calculated to be 0.028, which is significantly smaller than the coefficient obtained in the first experiment. A smaller coefficient of kinetic friction implies that there were less opposing forces on the iOlab in the second experiment, in comparison to the first. This may be caused by various differences between the two experiments, like the use of the wheels for the ramp experiment, or the gravity working for the iOlab when it was rolling down the ramp.
BRIDGING CONCEPTS: (answer 2 of the following) Go back to your Graphs from Method 2. Look closely at the time period where the iOLab was in free motion (after the initial tap). Note the acceleration seems constant, and is slightly negative; why is that? The value of the acceleration changes slightly at the top of the motion and is different coming back down than going up; why? Think about the Forces at play and what changes between the iOLab rolling up and down the ramp. Draw your free-body diagrams and include them in your report. As seen in figure, when the iOlab was in free motion the acceleration is observed to be constant, but also negative. This may be due to other inevitable resisting forces acting on the iOlab, such as friction, gravity, and air resistance. These forces oppose the motion of the iOlab, affecting its velocity. At the very top of the motion, the acceleration changes, this is caused by a change in direction. WHen the iOlab reaches the very top of its motion (finishes going up), it comes back down. The change of the iOlabs velocity causes a change in the acceleration. In addition, the acceleration also changes when the iOlab comes down the ramp to finish its motion. This is due to the outside forces acting on the iOlab. Although friction, and air resistance are still acting against the motion of the iOlab, gravity is now working with the motion of the iOlab, increasing its velocity in comparison to when it was going up the ramp. Method 2 made use of a ramp and component forces along the ramp to create equations to determine μ_ k. Would you be able to calculate the coefficient of friction between the wheels and a flat table (no ramp)? If possible, explain how you would set up your experiment, what devices you may want, and what iOLab sensors you would need to use. If not possible, give an explanation as to why. Does a negative coefficient of friction make sense? Explain why or why not using Newton's laws. What implication would/does this have in the real world? No, a negative coefficient of friction does not make sense, as it would suggest that friction is acting in the same direction as whatever force is applied. This is contrary to what friction actually is, a force that opposes motion. As Newton’s third law states, for every action, there is an equal and opposite reaction. Friction is the opposite reaction. The coefficient of friction is calculated by finding the ratio of the force of friction to the normal force, because these forces are both scalar, they do not have direction, only magnitude, therefore the ratio could not be negative.
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