Answer:
The amplitude of the subsequent oscillations is 13.3 cm
Explanation:
Given;
mass of the block, m = 1.25 kg
spring constant, k = 17 N/m
speed of the block, v = 49 cm/s = 0.49 m/s
To determine the amplitude of the oscillation.
Apply the principle of conservation of energy;
maximum kinetic energy of the stone when hit = maximum potential energy of spring when displaced
[tex]K.E_{max} = U_{max}\\\\\frac{1}{2} mv^2 = \frac{1}{2} kA^2\\\\where;\\\\A \ is \ the \ maximum \ displacement = amplitude \\\\mv^2 = kA^2\\\\A^2 = \frac{mv^2}{k} \\\\A = \sqrt{\frac{mv^2}{k}} \\\\A = \sqrt{\frac{1.25\ \times \ 0.49^2}{17}} \\\\A = 0.133 \ m\\\\A = 13.3 \ cm[/tex]
Therefore, the amplitude of the subsequent oscillations is 13.3 cm
