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A 1.25 kg block is attached to a spring with spring constant 17.0 N/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 49.0 cm/s . What are You may want to review (Pages 400 - 401) . Part A The amplitude of the subsequent oscillations

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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

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