This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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You're on a roll — Here are the solutions to your questions: b) The following results were obtained in an experiment to verify Hooke's law when a spring was extended by hanging various loads on it. (I) Complete the table for the extension e above. The original length of the spring (when Load = 0.00 N) is 10.00 cm. Extension is calculated as e = Length of spring - Original length. | Load (N) | Length of spring in cm | Extension (cm) | | :------- | :--------------------- | :--------------- | | 0.00 | 10.00 | 0.00 | | 1.00 | 11.50 | 1.50 | | 2.00 | 13.00 | 3.00 | | 3.00 | 14.50 | 4.50 | | 4.00 | 16.00 | 6.00 | | 5.00 | 18.00 | 8.00 | | 6.00 | 24.00 | 14.00 | (II) Plot a graph of load (y-axis) against extension. To plot the graph: • Draw a horizontal axis (x-axis) for Extension (cm) and a vertical axis (y-axis) for Load (N). • Choose appropriate scales for both axes to accommodate the data. For extension, values range from 0.00 cm to 14.00 cm. For load, values range from 0.00 N to 6.00 N. • Plot the points from the completed table: (0.00, 0.00), (1.50, 1.00), (3.00, 2.00), (4.50, 3.00), (6.00, 4.00), (8.00, 5.00), (14.00, 6.00). • Draw the best-fit line through the points. You will observe a straight line passing through the origin for the first few points (up to 4.00 N load), indicating Hooke's Law is obeyed. Beyond this point, the graph will curve, showing that the elastic limit has been exceeded. (III) From the graph determine the spring constant. The spring constant k is the gradient of the linear portion of the Load-Extension graph (F = ke k = (F)/(e)). We use the data points within the elastic limit (e.g., up to Load = 4.00 N). Step 1: Choose two points from the linear region of the graph. Let's use (e_1, F_1) = (0.00 cm, 0.00 N) and (e_2, F_2) = (6.00 cm, 4.00 N). Step 2: Calculate the gradient. k = (F_2 - F_1)/(e_2 - e_1) k = 4.00 N - 0.00 N6.00 cm - 0.00 cm k = 4.00 N6.00 cm k = (2)/(3) N/cm To express in SI units (N/m): k = (2)/(3) N/cm × 100 cm1 m k = (200)/(3) N/m ≈ 66.67 N/m The spring constant is 66.67 N/m. (IV) Calculate the energy stored when the spring is stretched to 16 cm. Assuming "stretched to 16 cm" refers to the total length of the spring. Step 1: Determine the extension. Original length, L_0 = 10.00 cm Final length, L = 16.00 cm Extension, e = L - L_0 = 16.00 cm - 10.00 cm = 6.00 cm. Convert extension to meters: e = 6.00 cm = 0.06 m. Step 2: Use the formula for elastic potential energy stored in a spring. E_p = (1)/(2)ke^2 Using k = (200)/(3) N/m and e = 0.06 m. E_p = (1)/(2) ((200)/(3) N/m) (0.06 m)^2 E_p = (1)/(2) ((200)/(3)) (0.0036) E_p = (100)/(3) × 0.0036 E_p = 100 × 0.0012 E_p = 0.12 J The energy stored when the spring is stretched to 16 cm is 0.12 J. Drop the next question.