Find the mass dropped onto the spring

[ehild]

My posts are almost identical with that of ILS.

ehild

 

[flyingpig]

No I am saying that ILS saying at the top of the equilibrium point, the change in height also happens to be d

 

[I like Serena]

No I am saying that ILS saying at the top of the equilibrium point, the change in height also happens to be d

To try to clarify I've drawn a picture:

Does this clarify it for you?

 

[flyingpig]

Then the height of the mass is d just before compression.

[PLAIN]http://img820.imageshack.us/img820/8761/fallingblockonspring.gif

 

[I like Serena]

"Just before compression" is the second picture.
The relative height is d above the point of full compression, which is in third picture.

 

[PeterO]

[PLAIN]http://img820.imageshack.us/img820/8761/fallingblockonspring.gif[/QUOTE]

The diagram was perfect BEFORE you wrote your d on it!

Go back to square one here.

Imagine you are going to perform the following experiment.

You are going to place the spring, normally found in the suspension of a car, on the floor. You are then going to drop a brick, from shoulder height, onto the spring.
That is what the problem is all about.

If I knew how tall you were, and how long the original spring was, I would be able to calculate exactly how fast the brick was traveling when it first contacted the spring. HOWEVER, I don't need those distances as I was already told the brick was traveling at speed v when it reached the spring - the calculation had already been done for me!

SO we only calculate from there on.

When the brick is finally stopped by the spring, it has traveled a further distance d.

It has lost mgd of potential energy since first contacting the spring. It has also lost all its kinetic energy: ½mv²

All that energy has been stored in the spring by compressing it an amount d.

SO

½kd² = mgd + ½mv²

re-arrange and solve for m.

Now the original question involved dropping something onto a spring, but I forget what, so I called it a brick.

NOTE: the object was DROPPED onto a spring, not placed on top of the spring.Peter

½ ⋅ ²