Room for Bml and John Rennie

Bml

5:54 AM

@JohnRennie Hi :-)

John Rennie

Hi :-)

Bml

Shall we see the problem?

John Rennie

Yes

yesterday, by Bml

On a platform of mass M = 20 kg, placed on a frictionless horizontal road, is a spring with elastic constant k = 1000 N/m and rest length L_0 = 0.5 m, which has one end attached to the platform and the other in contact with a mass M = 5 kg. There is friction between m and the platform, and the coefficient of kinetic friction is $\mu = 0.5$. Initially, the length of the spring is L = 0.2 m and it is released.
Find:
the velocities V_0 of the platform and v_0 of m when the spring reaches length L_0.

19 hours ago, by John Rennie

Bml

The initial energy of the system is 1/2 k ∆l^2 and the final energy is 1/2 m v_0² + 1/2 M V_0². Right?

John Rennie

Not quite ... you are not including the work done against the friction.

But your approach is fine.

Bml

6:04 AM

@JohnRennie This is the hot spot. My professor and I disagree on this point.

John Rennie

OK ...

I suspect your prof might be right on this one :-)

Bml

The work against the friction is - \mu mg ∆l, right?

John Rennie

Yes

Bml

The problem is that I consider ∆l negative (because there is a compression of the spring), so ∆l = 0.2-0.5 = 0.3, whereas my professor considered it positive (0.5-0.2 = 0.3)

So we arrive at two different numerical results...

Who is right and why?

John Rennie

It can be really complicated trying to keep track of all the signs with work, so I like to take a step back and think about where the energy is going.

We start with the energy stored in the spring:
E = ¹⁄₂kx²
and everyone agrees on that. Yes?

Bml

6:09 AM

@JohnRennie Yes

John Rennie

Now when we release the spring some of that energy gets converted to KE and some gets converted to heat by the friction:
¹⁄₂kx² = KE + heat

And both KE and heat are positive numbers.
Yes?

Bml

@JohnRennie Yes

John Rennie

So we are going to get:
¹⁄₂kx² = (¹⁄₂m₁v₁² + ¹⁄₂m₂v₂²) + FΔx

where the first term is the KE and the second term is the heat.
And both terms are positive.

Yes?

Bml

@JohnRennie Yes

John Rennie

So the heat term is going to be 25N × 0.3m

It can be confusing about exactly what signs to take, but at the end of the day we know the product must be a positive number.

Bml

6:21 AM

@JohnRennie But in this case I would be right, not my teacher...

John Rennie

I would say Δx is positive and F is positive, so I'd agree with your teacher.

Bml

@JohnRennie I would like to understand a little more about the issue because I am slightly confused

John Rennie

What direction are you taking to be positive?
I'm taking right to be positive since that's usual.

Bml

If I take W = Final Energy - Initial Energy, I have - \mu mg ∆l = ½ m v_0² + ½ M V_0² - ½ k ∆l², right?

@JohnRennie The spring is attached to a support placed on the right, so it moves to the left. I placed v_0 directed to the left and V_0 directed to the right...

John Rennie

So you are saying the final energy is:
½ m v₀² + ½ M V₀² + μmgΔ𝓁
Yes?

19 hours ago, by John Rennie

The way I've drawn it the support is at the left

Did you draw it the other way round?

Bml

6:30 AM

@JohnRennie Yes

John Rennie

Can we use my diagram since I don't want to redraw it?

Bml

@JohnRennie Yes, but isn't - \mu mg ∆l the work done by frictional force?

@JohnRennie Yes, obviously

First I used the theorem according to which W_{friction} = ∆E

John Rennie

@Bml No, μmgΔ𝓁 is the work done by the spring

Total work done by spring = work done by spring on masses + work done by spring against friction
Yes?

Bml

@JohnRennie And the one done by friction has a minus sign?

@JohnRennie OK

John Rennie

The force the spring is applying to the friction points right. Yes?

Bml

6:36 AM

@JohnRennie Yes

In your diagram, yes

John Rennie

And we are taking right right to be positive so the force the spring is applying against the friction is +μmg
Yes?

Bml

@JohnRennie Yes

John Rennie

And the displacement of the end of the spring is to the right, so the vector Δ𝓁 is to the right and also positive.
Yes?

Bml

@JohnRennie Wait: isn't the spring compressing?

John Rennie

No, we start with the spring compressed i.e. the initial state is with the spring compressed. Then we let go and the spring expands to the right pushing the block to the right.

Bml

6:42 AM

@JohnRennie Ah, OK, so yes, ∆l is positive

John Rennie

So the work done by the spring is +μmg × +Δ𝓁

Bml

@JohnRennie OK

So the final equation is...?

John Rennie

¹⁄₂kx² = ½mv₀² + ½MV₀² + μmgΔ𝓁

i.e. initial energy = final energy

Bml

@JohnRennie OK, so my professor was correct

Thank you :-)

John Rennie

You're welcome :-)

Bml

7:51 AM

@JohnRennie Hi... Do you remember this experiment?

John Rennie

Yes

Bml

Length at rest 8.2 cm

I have to attach different masses to the spring, right?

John Rennie

Yes, and for each different mass you measure the extension.

Bml

8:26 AM

Length at rest 8.3 cm, sorry

With 205.042 g, we have a length of 22.5 cm

With 49.920 g, length = 11.6 cm

With 25.077 g, length = 10.0 cm

With 19.970 g, length = 9.6 cm

With 9.992 g, length 8.9 cm

With 100.219 g, length = 15.3 cm

John Rennie

That looks pretty good.

Bml

@JohnRennie Wait: x is the final length, not the elongation...

John Rennie

It doesn't matter. The gradient will eb the same anyway.

And the y intercept will give our unloaded length.

Bml

@JohnRennie OK

So, I should do mg- k∆x = 0, right?

John Rennie

No, just use a linear regression

Bml

8:38 AM

@JohnRennie OK, but I should evaluated k

So I have to use ∆x = g/k m

John Rennie

This is what I get. The blue points are your data and the red line is the fit.

This is the link to the spreadsheet

Bml

@JohnRennie OK, so I have to equate g/k to gradient, yes?

@JohnRennie In this graph, if we put final lengths instead of elongations, the line does not pass through the origin. If we put elongations instead of final lengths, the line passes through the point (0,0), so it passes through the origin...

John Rennie

9:05 AM

In my fit the line passes through 8.2cm

So it's giving the relaxed length at 8.2cm whihc is very close to your measured value of 8.3cm.

Bml

9:25 AM

@JohnRennie OK, but if we put the elongation x instead of the final length, the first point will not be (0, 8.3), it will be (0,0), no?

John Rennie

Yes, but the gradient will still be the same.

Bml

OK, thanks

The documentation says: 1) determination of k_el from the relationship F = - k_el x and checking linearity... For what values of the elongations x of the springs is the linearity of Hooke's law not verifiable? Remark: the measurement of elongations is a relative measurement...

John Rennie

So you've done part (1) because the gradient is k/g so you've found k.

To do part (2) you need larger masses because the graph is linear for all the masses you used.

You need to keep increasing the mass until the points deviate from your straight line.

Bml

9:42 AM

@JohnRennie Isn't it g/k?

John Rennie

Oops yes.

mg = kx so x = (g/k) m

Bml

@JohnRennie OK

John Rennie

I think I was mentally graphing x on the x axis and m on the y axis.

Bml

@JohnRennie The fact is that we don't have larger masses in Lab...

And what does it mean "the measurement of elongations is a relative measurement..."?

John Rennie

Can you attach more than one mass?

If not then your conclusion will be that the extension remains linear up to the maximum mass available. That's a perfectly good conclusion.

Bml

9:50 AM

@JohnRennie I don't know

I don't understand "the measurement of elongations is a relative measurement..."

John Rennie

That statement does seem a bit odd.

Bml

10:33 AM

After this, they say: 2) Perform comparison between 3 equal springs: quality control.
3) Pendulum... with springs: T = 2π √(m/k_el)

John Rennie

I guess the point is that the three springs are supposed to be all the same, so they should have the same force constant.

So part 2 is just repeating what you did with two other springs of the same type to see how similar they are.

Then part 3 is setting the mass bouncing on the end of the spring and measuring the period.

So if you graph T² against m you should get a straight line plot.

Bml

11:08 AM

@JohnRennie I have to measure the period with a stopwatch?

Transcript for

Room for Bml and John Rennie