F - -ky EXPERIMENTAL PROCEDURE A. Determining the spring constant I, Hang the spring from a support, and suspend a weight hanger from the lower end. Add an appropri- ate weight to the hanger (for example, 100-200 g). nd record the total suspended weight [mig in new- 5 0608060 tons (N)] in Data Table J. Fix a meterstick vertically m alongside the weight hanger, and note the position of the bottom of the weight hanger on the meterstick. Slope - Record this as y, in the data table. 2. Add appropriate weights (for example, 100 g) to the weight hanger one at a time for incremental elonga- tions. Record the total suspended weight and the po- Elastic limit sition (in meters) of the bottom of the weight hanger on the meterstick after each elongation (y2. y's, etc.). The added weights should be small enough so that 7% alish ( Breaking point seven or eight weights can be added without nearing the spring's elastic limit. 3. On a sheet of graph paper, plot the total suspended weight force versus elongation position (mg versus y), and draw the straight line that best fits the data points. Determine the slope of the line and record (k, the spring constant). EXPERIMENT 13 / Potential Energy of a Spring B. Energy Considerations 1. Resuspend the spring with no mass, and measure and record the unstretched distance y. in Data Table 2. 4. Do this three times, and find an average yu. 5. Repeat steps 2-4 for my- 2. Hang mass m, on the spring and pull down 5-10 cm. Record this position as y2. 6. Using the obtained data and Eq. 13.3, determine if the potential energy stored in the spring is correct in 3. Release the mass, and estimate the highest position terms of the gravitational potential energy. (Does the ()) the mass reaches before reversing direction. equation balance?) k(V1 - Vo)2 - 2k(v2 - Vo)? = mgAy - Spring constant from the graph, k = 0.0207 N/m B. Energy Considerations Table 2. Determine potential energy yo = 0.00500 m my = 0.0015 kg m2 = 0.0025 kg yz (m) 0.0650 y1 (m) -0.0610 yz(m) 0.0650 y1 (m) -0.1020 y1 (m) -0.0592 y1 (m) -0.1010 yI (m) -0.0618 yI (m -0.1005 Avg. yi (m) -0.0607 Avg. y1 (m) -0.1012 For mass 1: Left side -5x 0.0207 x (0.0607 - 0.00500)2 - (0.0650 - 0.00500)?)(m?) = -5. 15 x 10-J Right side = migay = 0.0015 (kg) x 9.80 (2 x (0.0650 - (-0.0607))(m) = 6 32 x 10-$1