Spring constant for a block sliding PE and KE

[Zsmitty3]

1. An .8kg block is held in place against the spring by a 67 N horizontal external force. The external force is removed, and the block is projected with a v1=1.2. upon separation for the spring. The block then descends a ramp and has a velocity v2=1.9 m/s at the bottom. The track is frictionless up to that point. The block enters a rough section B, extending to E. The coefficient of friction over this section is .39. The velocity of the block is v3=1.4 m/s at C. The block moves on to D, where it stops. The spring constant of the spring is closest to:

1100N/m 3900N/m 2600N/m 1600N/m 2000N/m
2. It seems like a lot to take in. I have a couple thought about how to go about it but not sure if any would work. There's also a diagram of the situation included.
3. First I started simple and thought maybe we didn't need all that and tried just using F=.5kx^2 but we don't know x. So then I proceeded trying to use Conservation of Energy. At the top there would be some PE and the KE would be .576. after the drop, the PE would be 0 and the KE goes up to .5*.8*1.9^2=1.444. Since there is no friction, the initial PE must have been 1.444-.576=868N. That gets me to the spring with the PE and KE known but I have no clue from there because there's also PE in the spring along with the gravitational PE. I'm not sure if I need to include the friction part somehow. Thanks!

Homework Statement

 

[TSny]

First I started simple and thought maybe we didn't need all that and tried just using F=.5kx^2 but we don't know x.

Does .5kx² represent force or potential energy?

So then I proceeded trying to use Conservation of Energy.

If you know the KE of the block as it leaves the spring, what does that tell you about the initial PE of the spring?

 

[Zsmitty3]

Think I'm getting there but the x has me stuck

I see the correlation now. PE of held spring = KE of v1. So PE=.576 or .5kx^2=.576. We still have two variables though. Can I just solve for x and substitute? Or do I need to use the 67-N force that's holding the spring in place?

 

[TSny]

Two unknown variables: k and x. Need another equation. What is the relation between force and k and x?

 

[Zsmitty3]

Not sure? You mean like k-2F/x² or x= sqrt(2F/k)?

 

[TSny]

Did you cover "Hooke's law"? It relates the force of a spring to the spring constant k and displacement x.

 

[Zsmitty3]

Yes. F=1/2kx² you mean?

 

[Zsmitty3]

Oh wait no...that's F=kx...AHHHHHHH I see

So...maybe

PE=.576
.5kx²=.576

.5F²=.576

.5*49²... Oh wait nevermind

 

[TSny]

Yes. So...?

 

[Zsmitty3]

F=1/2mv²

F=.576

 

[TSny]

F=1/2mv²

F=.576

How can force equal kinetic energy? From PE = (1/2)kx² and F = kx you need to figure out how to get k. It's all come down to algebra.

 

[Zsmitty3]

How can force equal kinetic energy? From PE = (1/2)kx² and F = kx you need to figure out how to get k. It's all come down to algebra.

Bum bum bum. ALGEBRA. So wait maybe I was right and just stopped too soon.

so F = kx
so k=F/x

So 1/2(F/x)x²=PE

So 1/2 * Fx²/x

1/2* Fx =PE ?

 

[Zsmitty3]

No.x=F/k

1/2k(F/k)²

PE=1/2F²/k ?

Force if 67 so 1/2*67^2/k

I think I've just confused myself by working too many different problems...

 

[TSny]

Bum bum bum. ALGEBRA. So wait maybe I was right and just stopped too soon.

so F = kx
so k=F/x

So 1/2(F/x)x²=PE

So 1/2 * Fx²/x

1/2* Fx =PE ?

This is great. Can you use this to find x? Then you can find k.

 

[TSny]

No.


x=F/k

1/2k(F/k)²

PE=1/2F²/k ?

Force if 67 so 1/2*67^2/k

I think I've just confused myself by working too many different problems...

This will work too!

 

[Zsmitty3]

This will work too!

But what does this equal? PE. So 1/2*67²/k=PE

k= 67²/ 2*PE ? That doesn't make any sense to me though

 

[Zsmitty3]

Quote by Zsmitty3 View Post

Bum bum bum. ALGEBRA. So wait maybe I was right and just stopped too soon.

so F = kx
so k=F/x

So 1/2(F/x)x2=PE

So 1/2 * Fx2/x

1/2* Fx =PE ?

This is great. Can you use this to find x? Then you can find k.



So to solve for x. Fx=2PE. x=2PE/F
?

 

[TSny]

So to solve for x. Fx=2PE. x=2PE/F
?

Good, that's it. Then get k.

 

[Zsmitty3]

Not sure what to do from there though. Plug that value of x into 1/2kx² again? And set that equation equal to what? PE? or .576?

So

1/2K(2PE/F)²=PE ?
K*(2PE/F)²=2PE
K=(2PE)/(2PE/F)²?

 

[TSny]

What value did you get for x?

 

[TSny]

Not sure what to do from there though. Plug that value of x into 1/2kx² again? And set that equation equal to what? PE? or .576?

Yes, you could do that. But once you have x it might be easier to find k through F = kx.

 

[Zsmitty3]

Yes, you could do that. But once you have x it might be easier to find k through F = kx.

Oh.. Duhh

So k= F/(2PE/F)

k=F²/2PE or is my algebra wrong there?

 

[Zsmitty3]

Oh.. Duhh

So k= F/(2PE/F)

k=F²/2PE or is my algebra wrong there?

Well if that's correct

k= 67²/(2*.576)

k=3896.7 N/m or
k=~3900 N/m

 

[TSny]

I think that's right.

 

[Zsmitty3]

Well it's multiple choice and that's once so thanks for hanging in there with me :)