Spring (conservation energy) with friction

[neotriz]

Homework Statement

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

Homework 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.

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 don't know where to apply the Friction given.

 

[tiny-tim]

Welcome to PF!

Hi neotriz! Welcome to PF!

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?
…
PE[a]+KE[a]=PE+KE
…
But I don't know where to apply the Friction given.

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

so PE[a]+KE[a]=PE+KE + work done ​

 

[neotriz]

Hi neotriz! Welcome to PF!


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

so PE[a]+KE[a]=PE+KE + work done ​

Thanks

But I still don't get it.
If it that was the case, how would I find the velocity at ?

.

 

[neotriz]

Also, the answer must leave in cm unit. So I guess I have don't have to change the unit when it is compressing 2cm, right?

 

[tiny-tim]

If it that was the case, how would I find the velocity at ?

Just call it v

Also, the answer must leave in cm unit. So I guess I have don't have to change the unit when it is compressing 2cm, right?

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.

 

[neotriz]

Just call it v

I don't 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?

 

[tiny-tim]

Do we assume at x, PE is zero?

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

I don't 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?

Show us how far you've got with your calculation

 

[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

 

[tiny-tim]

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

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

and your s is related to x

 

[neotriz]

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

and your s is related to x

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?

 

[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)

 

[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

 

[tiny-tim]

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

 

[neotriz]

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

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

 

[tiny-tim]

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

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.