Conservation of Energy Question

[Arman777]

Homework Statement

there's a pic.

Homework Equations

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

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)
so we get quadratic equation.
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

 

[TomHart]

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

 

[Arman777]

I found 77 cm answer says 90 cm

 

[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.

 

[Arman777]

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.

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

 

[Arman777]

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

 

[haruspex]

$\frac 1 2kx^2-\frac 1 2mv^2=-F_{fr}x$ (here x are the same and represents the comprassion distance)

Did hitting the spring turn off gravity?

 

[Arman777]

Did hitting the spring turn off gravity?

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

 

[TomHart]

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

Me either.

 

[Arman777]

Did hitting the spring turn off gravity?

I found 0.93m when I did it.

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

Still having problem with (d)

 

[haruspex]

I found 0.93m when I did it.

Hmm... I think I got .90m.

Still having problem with (d)

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.

 

[Arman777]

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.

0.93 is approxemetly 0.90 I don't mind so much.

The answer is 15m for (d).

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

 

[gneill]

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.

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...

 

[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.

 

[haruspex]

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...

How silly of me.

 

[Arman777]

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

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

 

[TomHart]

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

 

[Arman777]

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

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

 

[TomHart]

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

 

[Arman777]

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

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

 

[TomHart]

Won't the spring end up compressed some because of the weight of the elevator when it finally comes to rest? The problem states that there is no friction when it comes to rest, so something has to be supporting the weight of the elevator.

 

[gneill]

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

Opinion is worth much less than calculation. What is the restoring force for a spring at its natural length? Will it hold up the cab?

 

[Arman777]

Yep I totally forget that.Ok thanks all of you

 

[haruspex]

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

Tom is right. Consider the forces on the cab when it has stopped.