Find the mass dropped onto the spring

[flyingpig]

Homework Statement

A block is dropped onto a spring with k. The block has a speed of v just before it strikes the spring. If the spring compresses an amount d before bringing the block to rest, what is the mass of the block?

The Attempt at a Solution

$$\int_{x_0 = 0}^{x_f = d} -kx\;dx + -mg(0 - d) = \Delta K$$

$$\frac{-kd^2}{2} + mgd = \frac{1}{2}mv^2$$

$$\frac{-kd^2}{2} = m\left(\frac{1}{2}v^2 - gd \right)$$

$$\frac{-kd^2}{v^2 - 2gd} = m$$

 

[flyingpig]

Wait v = 0 after because it comes to a rest.

So it should be

$$\frac{-kd^2}{-2gd} = \frac{kd}{2g} = m$$

 

[I like Serena]

Hmm, you seem to have made it difficult for yourself and there are a few signs wrong here.
Btw, your last remark is incorrect, because you have completely eliminated the initial speed v.
That can not be right, because the involved kinetic energy has to go somewhere.

Why don't you try it this way:

What is the total energy of the system just before the spring compresses?
And what is the total energy of the system when the spring is fully compressed?

These two need to be equal.

At any given time the total energy would be given by:
Energy = Gravitational energy + Kinetic energy + Spring energy

 

[flyingpig]

I thought there is no initial velocity because it was dropped?

 

[I like Serena]

I thought there is no initial velocity because it was dropped?

The height from which it was dropped is not given.
However, the speed just before compression is given...

 

[flyingpig]

Why can't you take intial velocity = 0 from when it was dropped? I thought these are path independent?

 

[I like Serena]

Why can't you take intial velocity = 0 from when it was dropped? I thought these are path independent?

Sure you can, but... what is the initial height?

 

[flyingpig]

I thought only the change in height mattered.

 

[I like Serena]

I thought only the change in height mattered.

Huh?
Mattered for what?

 

[flyingpig]

the energy calculations, why can't we say the change in KE is 0? I mean it dropped with 0 velocity and it came to a rest?

 

[I like Serena]

Please elaborate?

You can say that the change in gravitational energy is determined by the change in height.
That is correct.
What about it?

 

[flyingpig]

You asked me about the initial height. Sorry I was trying to say it didn't matter right?

 

[I like Serena]

We're not getting anywhere...

You don't know the initial height.
And it does matter.

What's the total energy at the various moments?

 

[flyingpig]

Well at the top it is mgh.

When it is dropped before touching the spring it is KE - PE = E?

 

[I like Serena]

Well at the top it is mgh.

Yes!

Note that v=0, so KE=0, and the spring is not compressed yet, so the spring-energy=0 as well.

When it is dropped before touching the spring it is KE - PE = E?

The proper formula would be KE + PE = E.

Where do you put your origin for the height?
What is KE and what is PE in this formula?

And what is the energy when the spring is fully compressed?
What is the formula for the energy of a compressed spring for that matter?

 

[flyingpig]

Oh I see, I assumed that i dropped it at the equilibrium point of the spring, but really it could have been dropped ABOVE the equilibrium point of the spring.

So the ball from whatever height had a "constant" speed of v when it is coming down and touching x = 0 the equilibrium point of the spring. In that case can i do this?

$$\int_{x_0 = 0}^{x_f = d} -kx\;dx + -mg(0 - d) = \Delta K$$

$$\int_{x_0 = 0}^{x_f = d} -kx\;dx + -mg(0 - d) = -\frac{1}{2}mv_0 ^2$$

EDIT: damn it doesn't work...

is this unsolvable?

 

[flyingpig]

Let me do this again

[PLAIN]http://img88.imageshack.us/img88/6199/unleduqh.jpg

$$\int_{x = 0}^{x = -x_0} -kx dx - mg(l -(|x_0| + h) - l) = \Delta K$$

$$\frac{-k}{2}(x_0^2) + mg[|x_0| + h|] = \frac{-1}{2}mv_0 ^2$$

Okay when it was still in the air and uncontacted with the spring

The following (so trivial!) took place

$$\frac{1}{2}mv_0^2 = mgh$$

$$h = \frac{v_0^2}{2g}$$

So subsitute everythign back in

$$\frac{-kx_0^2}{2} + mg[|x_0| + \frac{v_0^2}{2g}] = \frac{-1}{2}mv_0 ^2$$

Messed with a lot of algebra and eventually i got

$$\frac{kx_0^2}{2v_0^2 + 2g|x_0|} = m$$

If this is wrong, I am thinkiing of dropping out of physics and switching to pure math lo

EVERYTHING IS WRONG IN THIS POST

 

[flyingpig]

[PLAIN]http://img217.imageshack.us/img217/6425/unledjex.jpg

Reuploaded

The change in height is killing me

Initial height = l

Final height = d

l - (h + x₀) = d

[tex]\Delta height = (l - (h + x₀) - l) = - (h + x₀)

Does that look right??

 

[flyingpig]

Can somebody smart please login...

 

[flyingpig]

Sigh

$$l = d + |x_0| + \Delta h$$

Before contact of the spring and about to touch the spring's equilibrium point

$$mg(|x_0| + d - (\Delta h + |x_0| + d)) = \Delta K$$

$$-mg(\Delta h ) = \frac{1}{2}mv^2$$

I assume $$\Delta h < 0$$ because we are going down and we are losing potential? (got that idea from voltage lol)
$$E_i = mgl$$

$$E_f = \frac{1}{2}kx_0^2 + mgd$$

$$mgl = \frac{1}{2}kx_0^2 + mgd$$

$$m = \frac{kx_0^2}{v^2 + 2g|x_0|}$$

 

[flyingpig]

Are all the smart people asleep right now or what?

edit: contrary to belief I am more confident now in my answer $$m = \frac{kx_0^2}{v^2 + 2g|x_0|}|$$

 

[I like Serena]

Hi flyingpig!

I've been out of touch, sorry.
However, it seems to have paid off, since you got the right answer!

One thing to note though.
You seem to have mixed up x_o and d.
The problem states that d is the amount the spring compresses.
So in your final answer you should replace x_o with d.

 

[flyingpig]

I have two questions for you

1) Can you provide your solution? I found an answer way to do this problem afterwards and I want to see if it matches yours or you can provide a different method.

2) This was a simple problem right...?

 

[ehild]

Try with the conservation of energy. There are KE, gravitational potential energy and the elastic energy.

The object has some velocity v just before it touches the spring.
You can put the zero of potential energy there. What is the total energy?

The object compresses the spring by d and stops. What is the KE, potential energy and elastic energy in that position?

HE

 

[flyingpig]

Anyways the other way I did it was the following

$$\sum W = \Delta K$$

$$\int_{x = 0}^{x = x_0} -kx dx + mg(x_0) + mg(h) = -\frac{1}{2}mv_0^2$$

Basically I added up $$mg(x_0) + mg(h)$$ to mean that the total (positive) work done by the block as it falls. I didn't realize this was probably easier

 

[I like Serena]

I have two questions for you

1) Can you provide your solution? I found an answer way to do this problem afterwards and I want to see if it matches yours or you can provide a different method.

2) This was a simple problem right...?

All right, I think you have spent enough time and energy at this problem.


What you need to know it that total energy is given by:
E = kinetic energy + potential energy + elastic energy

which is:
E = (1/2)mv² + mgh + (1/2)kx²

Let's set the zero for height at the point where the spring is compressed.
Then the height of the mass is d just before compression.

Total energy just before compression is:
E = (1/2)mv² + mgd + 0

Total energy after compression is:
E = 0 + 0 + (1/2)kd²

Since total energy must be conserved, we have:
(1/2)mv² + mgd = (1/2)kd²

Divide left and right by everything except m and you have your answer.


And yes, with conservation of energy I'd call this a simple problem.

 

[flyingpig]

Let's set the zero for height at the point where the spring is compressed.
Then the height of the mass is d just before compression.

Why is it called d and the compression after also d?? How do you know they are going to be the same d?

 

[I like Serena]

Why is it called d and the compression after also d?? How do you know they are going to be the same d?

I'm not sure if I understand your question...

I'll just quote what the problem states (I didn't introduce d, the problem did):
"the spring compresses an amount d before bringing the block to rest"

So the change in height (h in the formula) is d, and the amount the spring compresses (x in the formula) is also d.

 

[flyingpig]

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

Isn't that assuming the height above the equilibrium point is the same as the compression?

 

[ehild]

The text of the problem says: "A block is dropped onto a spring with k. The block has a speed of v just before it strikes the spring. If the spring compresses an amount d before bringing the block to rest, what is the mass of the block?
At the beginning, the spring is unstretched, (its length is Lo) the top of the spring is somewhere, and you can put the zero of x there. The block touches the spring at x = 0.
The block moves downward compressing the spring by d. But the block is on the top end of the spring during its motion, so it goes down by d. The final position of the block is xf=-d.
The spring is compressed by d, so the change of length is ΔL=L-Lo = -d.

ehild

 

[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.