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So I have a block and a spring. The block is dropped onto a spring from a height. I know the mass of the block, how high from the spring it was dropped from, and the spring constant.

What I need to find out is the speed of the block, after the spring compresses a certain amount.

I know I can calculate the speed right before it hits the spring ($\sqrt{2ah}$ ), but how to find the reduced speed is eluding me. I know it has to do with work and/or energy because that is what we are doing in class right now.

Could someone please explain how to calculate the new speed?

edited Nov 30 at 2:23

asked Oct 27 at 16:52

Eric Koslow♦♦
118●1●9

Eric Koslow · Accepted Answer · 2009-11-30 02:25:04Z UTC

So after talking to my professor I have found the answer.

You know that the entire amount of energy in the system is the initial drop or mgh .

At the end, the energy in the system is represented by the kinetic energy of the block, plus the potential energy of the compressed spring, and subtracting the potential energy of the block falling past the point of reference (which in my case was the top of the uncompressed spring).

So the equation ended up being: $mgh = \frac{kx^2}{2}+\frac{mv^2}{2}-mgx$ | Where m is the mass of the block, g is the acceleration due from gravity, h is the initial drop height from the top of the uncompressed spring, k is the spring constant, v is the velocity of the block we are calculating for, and x is the amount the spring is compressed.

All one has to do now is plug in the numbers and solve for v.

How to find the speed of a block, after being dropped on a spring

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