Question

An elevator with mass 3000kg hangs still on a floor that's 5m above a spring(k=1,5*10^5). The cable of the elevator snaps and it falls with constant friction (4500N)

Answer 1

Hello! What is the question, here? It seems some has bee left out!

Answer 2

1)calculate the speed of the elevator before it hits the spring

Answer 3

2) The distance the spring will get pressed

Answer 4

3) The height the spring pushes the elevator up

Answer 5

it's alot..

Answer 6

Okay! Any idea how to start? Forewarning, I'll have to go in a little bit.

Answer 7

no idea, but i have an exam tomorrow so i'll have to check it tomorrow xD

Answer 8

So, how should we get started? Falling object because of gravity, opposed by friction... What's the net force?

Answer 9

it's own weight?

Answer 10

That, and the friction! This elevator is sqeezed in the walls, or maybe is on a pulley that has friction. Either way, the force of friction, \(F_f=4,500\ [N]\), as the problem states. And the force of weight is \(F_\text w=m_\text{elevator}\ g\) where \(g\) is the acceleration to gravity.

Answer 11

\(g\) is a negative number, for the record. What do you use, \(-9.8\ [m/s^2]\)?

Answer 12

yea

Answer 13

And the net force is \(F_\text{net}=F_\text w +F_f\).

So, shall we use substitution? \(F_\text{net}=m_\text{elevator}\ g+F_f=(3,000\ [kg])\ (-9.81\ [m/s^2])+4,500\ [N]\)

Answer 14

\(F_\text{net}=(3,000\ [kg])\ (-9.8\ [m/s^2])+4,500\ [N]\), sorry.

Answer 15

I got \(F_\text{net}=-24,900\ [N]\), but you should probably double check me...

Answer 16

i would, but i'll have to check tomorrow :P , doing this while studying for my exam :D

Answer 17

Now, it has a net force on it, because \(F_\text{net}\neq0\). So that acceleration gives it it's speed that we need to find after 5 meters.

Okay, good luck studying!

\(F_\text{net}=m_\text{elevator}\ a\), so \(a=\dfrac{F_\text{net}}{m_\text{elevator}}\)

And the velocity? \(v_f^2=v_i^2+2\ a\ d\implies v_f=\sqrt{v_i^2+2\ a\ d\ }\).

Answer 18

Then the elevator has a kinetic energy of \(E_k=\frac{1}{2}m\ v^2\).

To stop it, all that energy must be converted to (spring) potential energy as the elevator's momentum does work on the spring. The work on the spring is given by \(W=\frac{1}{2}k\ x^2\) I think. I have to go now..

But \(E_k=W\), and so you can solve for \(x\) to see how far the spring was compressed.

Answer 19

Thanks theEric ;)

Answer 20

Oh wait, I neglected friction.. Maybe I shouldn't have...

On the way up, the spring gives it velocity, but gravity accelerates it downwards. If you can, find out if there is still friction when it is on the spring! Good luck!

Answer 21

\(\sf\Large\color{orange}{Compressing\ the\ spring.}\)

So the kinetic energy that the elevator has when it hits the spring will be completely converted to spring potential energy and energies associated with friction (heat, sound, transferred kinetic...). So \(E_\text{k, before}=E_\text{P, after}+E_{f\text{, after}}\)

\(E_K=\frac{1}{2}m\ v^2\)
\(E_P= \frac{1}{2}k\ y^2\)
\(E_f=y\ F_f\)

So you can substitute them in to the above equation and solve for \(y\). It shouldn't be too tricky.

Answer 22

Oh, there is also the work done by the weight as the spring is compressed. So
\(E_\text{k, before}+E_w=E_\text{P, after}+E_{f\text{, after}}\)

\(E_w=m\ g\ y\), which makes it at least a little harder to solve for \(y\).

Answer 23

\(\sf\Large\color{lime}{Decompressing\ the\ spring.}\)

Now the energy is this:
\(E_{P\text{, spring}} +E_w+E_f=E_K\)
How high does the elevator go? Well, \(E_K=0\) when the elevator has stopped going up! Notice that, because of the direction of the forces, \(E_w\) and \(E_f\) will be negative. So set \(E_K\) equal to \(0\), and solve for \(y\) again.

Answer 24

I think that should be right... But I'm not 100% sure I got all of it right. But I hope it was valuable input, anyway! Best of luck!