# Conservation of Energy Question

1. Feb 1, 2017

### Arman777

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

2. Relevant equations
$W_{friction}=W_{fr}=Δ(ME)$
$U_{spring}=U_s=\frac 1 2kx^2$

3. The attempt at a solution
I found (a) using $W_{fr}=Δ(ME)$ which $v=7,4 \frac m s$
For (b) I wrote;

$\frac 1 2kx^2-\frac 1 2mv^2=W_{fr}$
$\frac 1 2kx^2-\frac 1 2mv^2=-F_{fr}x$ (here x are the same and represents the comprassion distance)
I found $\frac 1 2mv^2=48988J$
so $75000x^2+4400x-48988=0$
I solved it but didnt came the answer where did I go wrong ?

Thanks

2. Feb 1, 2017

### TomHart

What did you get for an answer? And what is the right answer?

3. Feb 1, 2017

### Arman777

I found 77 cm answer says 90 cm

4. Feb 1, 2017

### TomHart

Sorry, it took me so long to get back. I was off doing something else. I got 0.78 m (or 78 cm) for my answer.

5. Feb 1, 2017

### Arman777

then we get same results so there must be some roundings during the process...so I am correct I guess thank you

6. Feb 1, 2017

### Arman777

I solved (c) but (d) is hard.Can someone help me ?

7. Feb 1, 2017

### haruspex

Did hitting the spring turn off gravity?

8. Feb 1, 2017

### Arman777

Yeah...you are right.I didnt see that

9. Feb 1, 2017

### TomHart

Me either.

10. Feb 2, 2017

### Arman777

I found 0.93m when I did it.

I found also (c) correctly with an exact result.

Still having problem with (d)

11. Feb 2, 2017

### haruspex

Hmm... I think I got .90m.
d) seems a bit strange to me. Haven't you already worked out all the distances moved in b) and c)? Should just be a matter of adding them up. It feels like it is hinting at a shortcut to get this answer without using the intermediate ones, but I don't think there is one.

12. Feb 2, 2017

### Arman777

0.93 is approxemetly 0.90 I dont mind so much.

The answer is 15m for (d).

(b) is 0.9m and (c) is 2.8m

13. Feb 2, 2017

### Staff: Mentor

Consider the initial position and the eventual final (resting) position of the cab. The total energy input and eventually dissipated is determined by these points...

14. Feb 2, 2017

### Arman777

I guess I see...Last energy of our system is 47628J And he took 2.8m+0.9m=3.7m till now.He will go up and down until loses all energy which is 47628J.

15. Feb 2, 2017

### haruspex

How silly of me.

16. Feb 2, 2017

### Arman777

Then ;
$mgd-0=F_{fr}D$

here D comes 14.8m ?? I it this the answer

17. Feb 2, 2017

### TomHart

What are you using for the value of d? Are you including the displacement of the elevator to reach the equilibrium point?

18. Feb 2, 2017

### Arman777

d is in the given part.3.7 m.The block will stop there

19. Feb 2, 2017

### TomHart

Will it?
Disclaimer: I have been wrong so many times that I question my own judgment. :)

20. Feb 2, 2017

### Arman777

I think that way cause.That seems logical to me..lets wait for another opinion