# Elevator falling on spring question

1. Mar 13, 2012

### tal444

1. The problem statement, all variables and given/known data

Alright, this one's been bothering me.

An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable should break when the elevator is at a height h above the top of the spring, calculate the value that the spring stiffness constant k should have so that passengers undergo an acceleration of no more than 5.0g when brought to rest. Let M be the total mass of the elevator and passengers.

2. Relevant equations

W=$\frac{1}{2}$kd$^{2}$
W=Fd
E$_{p}$=Mgh

3. The attempt at a solution

I'm assuming that the potential energy will be equal to the kinetic energy, so:
W=Mgh=Fd

plugging in 5.0g for a I get h=5.0d, d=$\frac{h}{5.0}$

Mgh=$\frac{1}{2}$k($\frac{h}{5.0}$)$^{2}$
=$\frac{1}{2}$k($\frac{h^{2}}{25}$)
=$\frac{kh^{2}}{50}$
50Mgh=kh$^{2}$
k=$\frac{50Mg}{h}$

However, the answer in my textbook is $\frac{12Mg}{h}$. Any help here? I have a strange feeling that I did the first part wrong making the E$_{p}$ equal to E$_{k}$.

2. Mar 13, 2012

### tiny-tim

hi tal444!

i don't understand how you got d

use Ma = kd​

3. Mar 13, 2012

### tal444

I got d like this:
Mgh=Fd
Mgh=(M)(5.0g)d
d=$\frac{h}{5.0}$

and if I use Ma = $\frac{1}{2}$kd that gives me:

M(5.0g) = $\frac{1}{2}$kd
K = $\frac{10Mg}{d}$

which is still not the answer my textbook gives. Unless my textbook is wrong?

Last edited: Mar 13, 2012
4. Mar 14, 2012

### tiny-tim

hi tal444!

(just got up :zzz:)
ok, this is seriously wrong, let's go through it in detail …

i] Mgh is potential energy lost, so yes it has to equal kinetic energy gained = work done

and yes Fd = work done, but only if F is constant

if (as here) F isn't constant, you must use work done = ∫ F d(d)​

ii] (M)(5.0g)d is not a correct use of the information about 5.0g …

yes the maximum force is (M)0.5g, but ou have to equate that to the given spring force …

use (M)5.0g = kd

5. Mar 14, 2012

### tal444

I see, but this is only Grade 12 Physics, so I don't think we're supposed to use integrals yet. Is there any way to do the problem without integrals?

6. Mar 14, 2012

### tiny-tim

yes!!

use (M)5.0g = kd !

7. Mar 14, 2012

### tal444

Sorry if I sound like an idiot right now, my mind is pulling a blank. How am I supposed to find d using (M)5.0g = kd?

8. Mar 14, 2012

### tiny-tim

use the energy equation also

9. Mar 14, 2012

### tal444

Which energy equation, since I can't use the integral? I'm so lost xD

Last edited: Mar 14, 2012
10. Mar 15, 2012

### tal444

Alright, so I tried this:

Mg(h+d) = $\frac{1}{2}$kd$^{2}$
Mgh+Mgd = $\frac{1}{2}$kd$^{2}$
Mgh = $\frac{1}{2}$kd$^{2}$ - Mgd
Mgh = d($\frac{1}{2}$kd - Mg)
d = $\frac{Mgh}{0.5kd - mg}$

plugging in M(5.0g) for kd:

d = $\frac{Mgh}{2.5mg - mg}$
=$\frac{h}{1.5}$

putting this into d in M(5.0g) = kd:

k = $\frac{7.5Mg}{h}$

Still lost...

11. Mar 15, 2012

### tiny-tim

hi tal444!
well, that looks good

let's see … if we use M(6.0g) = kd instead of M(5.0g) = kd we get the right result

the question says …
passengers undergo an acceleration of no more than 5.0g when brought to rest​
… perhaps i'm being dim, but i still can't see any reason for adding an extra g

12. Mar 15, 2012

### tal444

Yeah, same here xD.

13. Mar 15, 2012

### emailanmol

Hello,

The answer in text book is right.

Try again :-)

Hint:

(M)(5g)=kd is wrong.

The basic principle is similar to what we will find for a falling block performing SHM

Last edited: Mar 15, 2012
14. Mar 15, 2012

### tal444

Sorry for sounding like an idiot again, but we haven't actually learned simple harmonics yet. Your hint is lost on me.

15. Mar 15, 2012

### emailanmol

Hey, no issue. i know what its like to be on the other side of doubts :-)

Think of it another way.
When the elevator falls on spring it compresses the spring (due to its force mg) .
As it compresses the spring a point x1 occurs where kx1=mg
i.e net force on spring is 0.
But any body stops when its velocity is 0, not its acceleration(This is one of the most important concepts in mechanics as most students make this mistake of stopping the body when its acceleration is 0).

So the point where the life stops is not x1 but is the point where its kinetic energy is 0.

Think of it this way,

At x1 the net force on object is 0.But its kinetic energy(or velocity) is not 0 .So the body overshoots beyond this point (from whereon againthe spring force acts decelerating it ) till the velocity is 0 .

This point x2 will also be the point where the particle faces its maximum acceleration(which you have to plug in as less than 5g).This point has max acceleration because maximum force acts on the particle here.( kx2-mg) .This is because x2 is the maximum compression in spring ( as its the last point where the body stops and thus the compresssion stops).

Use this to solve for k

Last edited: Mar 16, 2012
16. Mar 15, 2012

### tal444

So there is both the Mg from kx1 and then the M(5.0g) from kx2?
Therefore:

Mg + M(5.0g) = kd
6.0Mg = kd

plugging in M(6.0g) for kd:

d = $\frac{Mgh}{3mg - mg}$
=$\frac{h}{2}$

putting this into d in M(6.0g) = kd:

k = $\frac{12Mg}{h}$

Is that it? Sure hope so, cause this question is killing me.

17. Mar 16, 2012

### emailanmol

Yes,
Its right.

Now lets see why we got this answer so that all your doubts go away.

The ball starts falling from a height h above the top of spring.(we will measure all distances and potential energies wrt this point)
Till the point it reaches the spring due to action of gravitation, the body gains kinetic energy and loses its potential energy.
From here on another force acts on body which is the spring force.

As the body starts going down the spring compresses and provides an opposing force of value kx (and a potential energy corresponding to 1/2 kx^2 ) .
This force acts against the force if gravity .
A point x1 occurs where the acceleration of the body(which was facing downwards becomes 0) .At this point however the body still had speed and overshot (instead of stopping) the spring force increased further and the body's acceleration started pointing up.
This deceleration caused its speed to become 0.

At lowest point (which is the point where its kinetic energy is 0 as its speed is 0, because if speed is 0 the body wont go any further making it the lowest point)force acting is kx2-mg=ma.
Now we can see this acceleration will be maximum (as x2 is the lowest point corresponding to maximum value of x2 and thus max magnitude of acceleration)

We plug in the values
In the two the values into force and energy equation

Clear??

Last edited: Mar 16, 2012
18. Mar 16, 2012

### tal444

Indeed, thanks so much to both of you.

19. Mar 16, 2012

### Staff: Mentor

Suppose we consider the reverse scenario where the elevator begins at the bottom with the spring fully compressed, much like a spring based projectile launcher. In that case, in order for the passengers to actually undergo an acceleration of 5g, the net upward force must be

$f_{net} = M 5g$

But fnet is composed of the upward spring force and the downward gravitational force. Thus:

$f_{net} = F_s + F_g = k d - M*g$

so that

$F_s = M 5g + Mg = 6 M g$

This gives us

$k d = 6 M g~~~~or~~~~ d = \frac{6 M g}{k}$

The stored potential energy is then $\frac{1}{2}k d^2$. Energy conservation for when the "projectile" is at its maximum height h where KE is zero yields:

$\frac{1}{2} k d^2 - M g d - M g h = 0$

Use the expression for d and and solve for k.

20. Mar 16, 2012

### tiny-tim

hi gneill!
thanks!

(i thought of doing that, but somehow i got confused, and made it 4Mg instead of 6Mg )