# Homework Help: Spring (conservation energy) with friction

1. Mar 15, 2009

### neotriz

1. The problem statement, all variables and given/known data
A block of mass 1.6 kg is attached to a horizontal spring that has a force constant 900 N/m. The spring is compressed 2.0 cm and is then released from rest.

a) A constant friction force of 3.8 N retards the block's motion from the moment it is released. At what position x of the block is its speed a maximum?

m=1.6kg
K=900 N/m
Friction=3.8N
s=2cm

2. Relevant equations

PE=.5Kx^2 (elastic potential energy function from the spring)
KE=.5mv^2
PE[a]+KE[a]=PE+KE, where [a] is the initial position of the question, and is where the position of x.

3. The attempt at a solution

Since [a] is the initial position, then the equation that I wrote above will get rid of KE[a], since there is no Kinetic energy involved (It is at rest). Furthermore, at , it will get rid of PE because when finding the speed is max, PE is zero and KE is at the max.

That being said, what throws me off is the friction that slows down the block after it is released.

So my next guess was since we got rid of KE[a] and PE, I put the equations accordingly like this:

.5Kx^2=.5mv^2

But I dont know where to apply the Friction given.

2. Mar 15, 2009

### tiny-tim

Welcome to PF!

Hi neotriz! Welcome to PF!

The work-energy theorem states that work done = energy lost,

so PE[a]+KE[a]=PE+KE + work done

3. Mar 15, 2009

### neotriz

Re: Welcome to PF!

Thanks

But I still dont get it.
If it that was the case, how would I find the velocity at ?

.

4. Mar 15, 2009

### neotriz

Also, the answer must leave in cm unit. So I guess I have dont have to change the unit when it is compressing 2cm, right?

5. Mar 15, 2009

### tiny-tim

Just call it v
No, wrong!

Change into m for formulas like 1/2 kx2, and change back again at the end, or the squares can get you into serious trouble.

6. Mar 15, 2009

### neotriz

I dont understand

When I do the calculation, I have everything except the displacements (the one it is asking me to find) and the max velocity. How would I find the latter one?

One more thing: Do we assume at x, PE is zero?

7. Mar 15, 2009

### tiny-tim

You can set PE = 0 anywhere … PE is always relative

but it makes the maths a lot easier if you set it at 0 at the uncompressed length
Show us how far you've got with your calculation

8. Mar 15, 2009

### neotriz

Here's what I have:

PE[a]=KE+Work done by friction (correct)?

thus

.5K(x)^2 = .5mv^2 + Work done

.5(900N/m)(2.00 X 1-^-2 m)^2 = .5(1.6)v^2 + 3.8N(s)

NOTE: I changed 2.00 cm to 2.00 X 10^-2

Through alebraic manipulations, I get:

.18 = .8v^2 + 3.8(s), where I guess (s) is what I am trying to find

9. Mar 15, 2009

### tiny-tim

No, you need .5Kx2 (with different x) on both sides

and your s is related to x

10. Mar 15, 2009

### neotriz

So that means
what I wrote before:
PE[a]=KE+Work done

"The Work Done" is included the Friction's and the Spring's at x?

11. Mar 15, 2009

### tiny-tim

perhaps I should have been clearer …

the work done in that equation is only the work done by friction

(technically, PE is work done, but when it's a conservative force, such as gravity or a spring, we call it PE instead)

12. Mar 15, 2009

### neotriz

So where is the other .5Kx^2 from the otherside is coming from?

I'm really sorry, I'm new at this physics thinking

13. Mar 15, 2009

### tiny-tim

One side is the initial energy of the spring, the other side is the energy of the spring at a general time

14. Mar 15, 2009

### neotriz

But I thought at x, there is no PE, meaning KE is at max.

15. Mar 16, 2009

### tiny-tim

uhh?

x is at max KE, so at min PE, not zero PE

you can set PE = 0 at any position you like

but if you want to use PE = 1/2 kx2, then x is measured relative to the uncompressed length, and the PE will therefore be 0 at the uncompressed length.