A basket of negligible weight hangs from a vertical spring scale of force constant 1500 N/m. (a) If you suddenly put a 3.0-kg adobe brick in the basket, find the maximum distance that the spring will stretch. (b) If, instead, you release the brick from 1.0 m above the basket, by how much will the spring stretch at its maximum elongation?
Solution 74P Step 1: a) The force due to gravitational force on the basket will be, F = mg Where, m - mass of the brick g - acceleration due to gravity Provided, m = 3.0 kg and take g = 9.8 m/s 2 F = 3.0 kg × 9.8 m/s = 29.4 N Then, we can write, F = 29.4 N (The force is due to the gravitational field) spring But we know that, F = kx x = F/ k Where, k - force constant of the spring and x - elongation of the spring. Provided, k = 1500 N/m x = 29.4 N/ 1500 N/ m = 0.0196 m = 1.96 cm Elongation of the spring, x = 1.96 cm
