# Falling objects, springs

1. Sep 11, 2004

### aliendoom

hi,

how can I figure out the peak velocity of 1kg dropped 1m onto a spring? i know the peak velocity occurs when F = -kx + mg = 0.

Last edited: Sep 11, 2004
2. Sep 11, 2004

### Integral

Staff Emeritus
The peak velocity will occur when the mass first reaches the spring after its fall.

3. Sep 11, 2004

### aliendoom

you need to rethink that.

Last edited: Sep 11, 2004
4. Sep 11, 2004

### Tide

The maximum speed will occur when the gravitational force is balanced by the spring force: $x = -\frac{mg}{k}$
Before that point is reached the object is still accelerating and beyond that point the acceleration is upward.

5. Sep 11, 2004

### aliendoom

thanks. anyone else?

6. Sep 11, 2004

### Integral

Staff Emeritus
The instant the mass hits the spring it begins to decelerate. Perhaps we are not speaking of the same speed. I am referring to the mass, and you?

7. Sep 11, 2004

### Tide

No, because the spring force is zero at that point and gravity is still acting.

8. Sep 11, 2004

### aliendoom

in my initial post i tried to indicate how far i had gotten. i can get x. i can also get the velocity before the spring starts stretching. what i need is the peak velocity. that's the part i'm having trouble with.

Last edited: Sep 11, 2004
9. Sep 11, 2004

### Tide

Use energy conservation!

10. Sep 11, 2004

### Integral

Staff Emeritus
Ok, ok...
Looks to me like once we hit the spring we must satisfy the DE

$$\ddot {x}m = -mg - kx$$

Find the velocity expression from this and you are done. :)

I'm workin on it.

Edit: inserted a lost factor.

Last edited: Sep 16, 2004
11. Sep 11, 2004

### aliendoom

how? the spring continues to stretch after kx = mg.

12. Sep 11, 2004

### Tide

No, the spring is being compressed. Remember, you're DROPPING the object onto the spring.

Do you know what the potential energy is for a spring and how to calculate gravitational potential energy?

13. Sep 11, 2004

### aliendoom

a flicker of a light bulb. 1/2kx^2 is the potential energy of the spring, so subtracting that from the change in potential energy of the falling mass should give me the kinetic energy. am i right?

Last edited: Sep 11, 2004
14. Sep 11, 2004

### Tide

That's part of it. Energy conservation tells you that
$$mv^2 + kx^2 + 2mgx = mv_0^2$$
is a constant. I've taken the zero point of the potential to be at the point of contact.

Last edited by a moderator: Sep 16, 2004
15. Sep 11, 2004

### Integral

Staff Emeritus
From the solution of the DE I get
Vi = velocity at the top of the spring.
$$x(t) = \frac {mg} k Cos ( \sqrt { \frac k m}t) + V_i \sqrt {\frac m k} Sin (\sqrt {\frac k m } t) -\frac {mg} k$$
The velocity equation is:

$$\dot {x}(t) =\sqrt {\frac m k}g Sin(\sqrt{\frac k m}t) + V_i Cos{(\sqrt {\frac k m}t)}$$
The time at which the max velocity occurs will be

$$t_m =\sqrt {\frac m k} Tan^{-1} (\frac g {V_i} \sqrt { \frac m k})$$

Simply plug this time into the velocity equation to find the Max velocity.

Edit the edit: All units are now happy!

Last edited: Sep 16, 2004
16. Sep 11, 2004

### Tide

17. Sep 11, 2004

### Integral

Staff Emeritus
Thats for sure!

Got 'em fixed now. A lost constant or 2 can really mess things up!

18. Sep 11, 2004

### aliendoom

ok but shouldn't it be:

$$KE_{mass} + PE_{spring} = \Delta PE_{mass}$$

when $$v_{0}=0$$.

so,

$$1/2mv^2 + 1/2kx^2 = mg(h+x)$$

or

$$mv^2 + kx^2 - 2mg(h+x) = 0$$

where h is the drop distance and x is the spring compression.

in other words in your equation the distance in the $$kx^2$$ term shouldn't be the same as the distance in the $$2mgx$$ term, and it seems to me the sign of the $$2mg(h+x)$$ term should be negative. as far as i know, energy isn't a vector quantity, so i'm not sure how this would result in different signs:

also, these equations seem to express the same thing I said in my previous post:

(how do you get two lines of equations between a set of tags? i tried \\ at the end of a line and the equations just ran together.)

Last edited by a moderator: Sep 16, 2004
19. Sep 11, 2004

### Tide

The full equation would be
$$mv^2 + kx^2 + 2mgx = mv_0^2 + kx_0^2 + 2 mgx_0$$
where the subscripts refer to some initial value. I took $x_0$ to be zero which corresponds to the point of initial contact. My $v_0$ is the speed of the object just as it hits the spring which you can calculate if you know the height above the spring when it was released.

Last edited by a moderator: Sep 16, 2004
20. Sep 12, 2004

### ehild

$$mv^2 + kx^2 -2mgx = mv_0^2$$
V is maximum when
$$x=\frac{mg}{k}\mbox{. } v_{max}=\sqrt{v_0^2+\frac{mg^2}{k}}.$$

ehild

Last edited by a moderator: Sep 16, 2004