Posted by **Diane** on Sunday, November 5, 2006 at 9:27am.

I have a question involving the spring costant:

A 2-kg block is attached to a horizontal ideal spring with a spring constant of 200N/m. When the spring has its equilibrium length the block is given a speed of 5 m/s. What is the maximum elongation of the spring? The answer is supposed to be in m.

I used 1/2mv^2=1/2KX^2, the answer I get is .5m which doesn't seem right.

Thank you

The word equilibrium is confusing here, so I am assuming as you did that the equilibrium length for a horizontal spring is its stretched length.

You used the correct relationship: max PE equals starting KE.

then if follows

x= v/sqrt k

I dont get your answer, I get less.

did u get .05m? i don't why i am getting that now

## Answer this Question

## Related Questions

- physics - have a question involving the spring costant: A 2-kg block is attached...
- Physics - Spring - Suppose a 1.5kg block of wood is slid along a floor and it ...
- physics - A block of ice of mass 4.10 kg is placed against a horizontal spring ...
- physics - A 395.0 g block is dropped onto a vertical spring with a spring ...
- physics - A 395.0 g block is dropped onto a vertical spring with a spring ...
- Physics - A 395.0 g block is dropped onto a vertical spring with a spring ...
- physics - A 390.0 g block is dropped onto a vertical spring with a spring ...
- Physics - please help me.. - A 395.0 g block is dropped onto a vertical spring ...
- Physics - please help!! - A 395.0 g block is dropped onto a vertical spring with...
- physics - A 4 kg block is pushed by an external force against a spring with ...

More Related Questions