A 0.164-kg block is dropped straight downwards onto a vertical spring. The spring constant of the spring is 62 N/m. The block sticks to the spring and the spring compresses 0.17 m before coming to a momentary halt. What is the speed of the block just before it hits the spring? Give your answer in (m/s).(adsbygoogle = window.adsbygoogle || []).push({});

My initial thought was to use a conservation of energy formula. I came up with:

(1/2)(k)(x)^2 = (1/2)(m)(v)^2

(1/2)(62 N/m)(0.17 m)^2 = (1/2)(0.164 kg)(v)^2

v = {[(62 N/m)(0.17 m)^2]/(0.164 kg)}^(1/2)

This, however, is not the correct answer. Therefore, there must be another way of doing this problem. Can anyone help me?

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# The Speed of a Block (Springs)

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