Physics 131- Section L Thursdays 9:00 AM ILC S110 June 29, 2017 Work Theorem at Work Abstract This lab highighted the use of Kinetic energy and the Work energy theorem. A hanging weight was attached to the PAScar on an almost frictionless track. The weights varied in size from 5g, 10g and 20g. A sensor captured the velocity and position of the car. This data was used to calculate displacement and acceleration over the course of time. Questions & Answers 1. Consider the PAScar and hanging and mass system. The PAScar has a mass of 250 grams (0.250kg) and hanging mass weighs m=25grams (0.025kg) descends starting at rest a total distance of 1m. At the half way mark (0.5m) the string breaks. What is the kinetic energy of the PAScar at the moment the string breaks? (2 points) The kinetic energy at the time the string breaks is 0.11135 J. Figure 1 (below) shows these calculations. Figure 1 shows the calculation of the kinetic energy at the time the string breaks2. Plot the v2 vs x-xi graph for each of your three sets of measurements. Perform a linear fit for each graph and determine the slope of best fit. (3.0 points) Graph was created using data points of (x-xi) and v^2 for different hanging masses. The linear fit was done within excel, and the best fit slope is shown on each graph within the linear equation. Figure 2: Displacement vs. Acceleration for a 5 gram hanging mass Slope of best fit: y=0.2589x-.0002 Table 1: Table values used to create Figure 1 (x-xi, v^2)Figure 3: Displacement vs. Acceleration for a 10 gram hanging mass Slope of best fit: y=0.6027x +0.0012 Table 2: Table values used to format Figure 3 (x-xi, v^2)Figure 4: Displacement vs. Acceleration for a 20 gram hanging mass. Slope of best fit: y=1.29x -.0011 Table 3: Table values used to format Figure 3 (x-xi, v^2) 3. The slope of best fit equals 2mg/(m+M) For each value of slope, m, and M calculate the value of g and compare them to the expected value. What is the percent difference for each measurement? (3.0 points) By rewriting the formula, we can make it solve for g. Giving us, (slope(m+M))/(2m)=g . M was always 250g, due to the unchanging mass of the PAScar. Variable m was the mass, in grams, of the hanging weight, which varied between trials. The slope inputted was from the linear fit equation of each graph, (.2589, .6027, 1.29) respectively for the 5 gram, 10 gram, and 20 gram hanging weight. Step by step calculations shown below. Hanging mass (grams) Calculated g (m/s^2) Percent error (%) using 9.8 m/s^2 as expected g. ((Calculated g- expected g)/expected g)*100 5 6.60 32.65% 10 7.84 20% 20 8.71 11.12%4. A mass m=0.2 kg is hung from a string. The other end of the string is attached to a PAScar of mass M=250 grams. The PAScar and hanging mass start from rest (vi=0) and the hanging mass descends a distance of 1 m. Using Eq. 3.4 from the manual, calculate the velocity, vf , of the PAScar. (2.0 point). The vf of the PAScar is 2.95 m/s. This was done in a similar process to question one. Handwritten calculations of vf are below.5. Let’s consider the system in reverse. The PAScar (M=250 g) is moving up the track with an initial velocity (vi) until it stops (vf=0). What must the initial velocity be in order to lift a hanging mass m=0.15 kg on the string a height of 1 m. (2.0 point). The initial velocity must be 2.711 m/s. Based on calculations below. 6. For the same amount of work done to move the PAScar, if the mass of the PAScar increased, would the final velocity vf increase, decrease, or stay the same? Explain why using conservation of energy concepts. (2.0 points) The final velocity would decrease. Using the equation vf= Sqrt((2*w)/M)) where M is the mass of the PAScar. If w is assumed constant, than an increase in M is an increase in the denominator will result in a lower final velocity. Work is a measure of the transfer of energy by aforce acting on a displaced object. I believe an object with greater mass having greater inertia, and therefore requiring more energy to move. Applying this logic, if you had two objects one with a larger mass and one with a smaller one, the object with larger mass will move will a velocity less than that of the smaller mass. 7. It is not actually possible for us to completely isolate the PAScar and hanging mass system. Name at least one force that we didn’t account for during this experiment that would affect conservation of energy. (1.0 point) We did not account for the magnets within the PAScar. While the magnetic force is negligible until the PAScar is close to the magnet, it is still constantly acting on it which could cause some skew in data. 8. When we “lose” energy in the form of dissipated energy, where does it go? (1.0 point) Energy is never lost but it is only transferred into another form. One way that energy can be dissipated is in the form of heat, meaning it won’t appear in the form of kinetic energy and will appear “loss” but is actually still in the system. Conclusion Energy is a force involved in systems that can appear in many forms, even several over the course of one experiment.. An understanding of the work energy theorem and kinetic energy allow for further expansions on topics we have learned thus far, such as tension and F=m*g. This experiment allowed for us to understand the variable relationships within the