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## Homework Statement:

- A 2 kilogram block is dropped from a height of 0.45 meter above an uncompressed spring. The spring has an elastic constant of 150 newtons per meter and negligible mass. The block strikes the end of the spring and sticks to it. Determine the maximum compression of the spring.

## Relevant Equations:

- U = 1/2kx^2, K = 1/2mv^2, U = mgh, F = kx

I found the amplitude of the simple harmonic motion that results (0.367, and I know this is correct because I entered it and it was marked as a correct answer), and assumed it would be the same value for the maximum compression since x(t) = Acos(wt). And, since the maximum value of cosine is 1, then the maximum value of x(t) would be A. But it's not correct. I also tried doubling this value, since I think the amplitude is the distance from the equilibrium position...so I'd double it to find the total compression. This is still incorrect.

Then I tried conservation of energy. I know the block has a potential energy of mgh, or 2 x 9.8 x 0.45. I made that equal to the energy of a spring, which is 1/2 x 150(x^2). Setting both equal to each other, I get 0.343, which is also incorrect.

Please help! I don't know what other approach to take for this question.

Then I tried conservation of energy. I know the block has a potential energy of mgh, or 2 x 9.8 x 0.45. I made that equal to the energy of a spring, which is 1/2 x 150(x^2). Setting both equal to each other, I get 0.343, which is also incorrect.

Please help! I don't know what other approach to take for this question.

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