Understanding why the initial height of one block is equal t

[RoboNerd]

Homework Statement

A 20.0-kg block is connected to a 30.0-kg block by a string that passes over a light frictionless pulley. The 30.0-kg block is connected to a spring that has negligible mass and a force constant of 250 N/m. The spring is unstretched when the system is as shown in the figure, and the incline is frictionless. The 20.0-kg block is pulled20.0 cm down the incline (so that the 30.0-kg block is 40.0 cm above the floor) and released from rest. Find the speed of each block when the 30.0-kg block is 20.0 cm above the floor (that is, when the spring is unstretched.

The problem is on page 62 of the document on the attached website along with the diagram. [/B]
http://www.santarosa.edu/~lwillia2/40/40ch1011_f14.pdf

Homework Equations

: [/B]
Conservation of mechanical energy. Mechanical Energy initial = Mechanical Energy final

The Attempt at a Solution

I prove that I attempted to solve the problem by attaching a picture below of my unfinished sheet.
So my high school teacher assigned the problem above. I just took the link from a college website because it has the problem with diagram and the explanation. I also have the following photographs from the student solutions manual that the book uses, which are attached below.

I understand how conservation of mechanical energy works in this case, but I am struggling with another aspect of the problem.
So I am struggling to understand one critical aspect of their explanation... why does the height of the 20.0 kg block when it is pulled back equal the height of the 30.0 kg block when the spring becomes unstreched?

I would have thought that because there is a 40 degree angle that things would have been different...

Could someone please explain?

Thanks in advance!

 

[Simon Bridge]

Clearly the height of the block depends on the length of the string connecting them together (and, yes, the angle of the ramp).
Do they tell you the length of the string? Does the length matter for the thing you have to find out?

 

[RoboNerd]

Clearly the height of the block depends on the length of the string connecting them together (and, yes, the angle of the ramp).
Do they tell you the length of the string? Does the length matter for the thing you have to find out?

Nothing was said about the string in any way shape or form.

 

[faradayscat]

When they say "The 20.0-kg block is pulled 20.0 cm down the incline" they mean the vertical height, not the actual length of the ramp. Indeed the length down the ramp is:

$L = \frac{h}{sinθ}$

 

[RoboNerd]

When they say "The 20.0-kg block is pulled 20.0 cm down the incline" they mean the vertical height, not the actual length of the ramp. Indeed the length down the ramp is:

$L = \frac{h}{sinθ}$

So how does this prove that the initial position of the 20.0 kg block is equal to the final position of the 30.0 kg block?

 

[WrongMan]

The block on the incline (lets call it block 1) is pulled down 20 Cm.. which (coincidentally) is the same height that the other block is when the system is at rest.
there is no logical explanation as to why the block1 is pulled down 20 cm, its just the way the exercise is set, if the block1 were to be pulled down 50 cm the rest position for the system (or unstreached position for the spring) would still be with block2 at 20cm

now, if you meant, why is it that when you pull block1 20 cm, the other block also rises 20 cm, then that's just logic, 20cm is 20cm in whatever direction you mesure it...the distance block1 travels is exacty the same distance block2 travels, just in diferent direction

 

[RoboNerd]

The block on the incline (lets call it block 1) is pulled down 20 Cm.. which (coincidentally) is the same height that the other block is when the system is at rest.
there is no logical explanation as to why the block1 is pulled down 20 cm, is just the way the exercise is set, if the block1 were to be pulled down 50 cm the rest position for the system would still be with block2 at 20cm

now, if you meant, why is it that when you pull block1 20 cm, the other block also rises 20 cm, then that's just logic, 20cm is 20cm in whatever direction you mesure it...if the distance block1 travels is exacty the same distance block2 travels, just in diferent direction

No, I am not asking that. I am told that the vertical height of the block that is pulled 20 cm at the point where it is pulled away is equal to the vertical height of the equilibrium position of the 30 kg block at equilibrium...

so I have the following... as shown in my photograph.

Why is this the case?

 

[faradayscat]

It's just a coincidence, it just means the 20 kg block was at an initial height of 40 cm.

 

[RoboNerd]

It's just a coincidence, it just means the 20 kg block was at an initial height of 40 cm.

But they do not give me the initial height so logically, I can not even derive the fact that the 20 kg block was at an initial height of 40 cm.
And thus, I can not logically derive what the solutions establish to be fact: that the height of the 20 kg block when it is pulled back equals that of the 30 kg block at equilibrium height.

... so how could I solve the problem without realizing (or making up) this "coincidence?"

 

[WrongMan]

No, I am not asking that. I am told that the vertical height of the block that is pulled 20 cm at the point where it is pulled away is equal to the vertical height of the equilibrium position of the 30 kg block at equilibrium...

so I have the following... as shown in my photograph.

Why is this the case?

I don't see anywhere that says that the block1 when pulled 20cm along the incline ends up at an height of 20cm relative to the horizontal... even if it does, that is just a pure coincidence... it certainly doesn't say that (in relation to the horizontal) its original position had an height of 40cm... that just means his original height was 20+20sen(θ)

PS: i can't read the solutions photograph

 

[Simon Bridge]

Nothing was said about the string in any way shape or form.

What's happening is prob that the thing you are trying to find does not depend on the length of the string (and thus the height of the block) ... so you are free to pick any length you like for that string.
The best thing to do is to pick a length that makes the math simpler for you.
The authors chose the length so that the height of the black on the ramp is the same as the height of the other one when the spring in uncompressed.
You want to see how that changes things - try using a different variable for the height of the block on the ramp.

[read other replies - and maybe they have it right instead]

 

[faradayscat]

You don't need this information, all you know is that the height of the 20 kg increases decreases by 20 cm and the height of the 30 kg increases by 20 cm, that is enough to solve the problem.

 

[RoboNerd]

What's happening is that the thing you are trying to find does not depend on the length of the string (and thus the height of the block) ... so you are free to pick any length you like for that string.
The best thing to do is to pick a length that makes the math simpler for you.
The authors chose the length so that the height of the black on the ramp is the same as the height of the other one when the spring in uncompressed.
You want to see how that changes things - try using a different variable for the height of the block on the ramp.

A friend of mine said he was able to solve the problem by adjusting various string lengths, but that seems a little unsystematic to me... Would such an approach work for sure?

 

[RoboNerd]

Ohh...

I am attaching a new photo of the solutions manual here. Hope this helps

 

[RoboNerd]

A friend of mine said that he was able to solve the problem by adjusting rope lengths, but this seems a little unsystematic... and I am hesistant to do this... plus adding in a string length of my own choosing can mess up the problem, can it not?

 

[Simon Bridge]

A friend of mine said he was able to solve the problem by adjusting various string lengths, but that seems a little unsystematic to me... Would such an approach work for sure?

It doesn't matter what the string length is ... the only reason to fiddle with it is to make some numbers cancel out.
It is better practise to assign the string length to an unknown variable, say "s".
In general - do all the algebra in symbol form and only substitute the numbers later.

 

[WrongMan]

Its a solutions typo, where they switched the masses, the pdf file you inserted has the correct form... it should be 20*0.2*sen(θ)*g and 30*g*0.2 not the other way around

 

[RoboNerd]

It doesn't matter what the string length is ... the only reason to fiddle with it is to make some numbers cancel out.
It is better practise to assign the string length to an unknown variable, say "s".
In general - do all the algebra in symbol form and only substitute the numbers later.

Good idea... I definitely have to try that.

Question: If I have my string length "s"... then how would I represent the length of the string parallel to the incline and the length of the string hanging vertically?

 

[RoboNerd]

WrongMan, what line of the solutions manual are you referring to?

 

[WrongMan]

WrongMan, what line of the solutions manual are you referring to?

the part where they "particularize to the problem situation" and the part where they substitute for values.
they switched the masses. where m1 is there should be an m2, where m2 is there should be an m1.

or not I am squinting to much

 

[RoboNerd]

the part where they "particularize to the problem situation" and the part where they substitute for values.
they switched the masses. where m1 is there should be an m2, where m2 is there should be an m1.

or not I am squinting to much

They define the 30.0 kg mass to be mass 1... do you also have this in mind?

 

[WrongMan]

forget about it, the solutions are alright.
what they do is they set an "imagined" length of string you don't need to know it, they just set the initial height of the 20kg mass (initial as in when it is pulled) to be the 0 of the referential, (and make it so its actual height is 20cm), so that the final potential energy of the 20kg mass is 0.2(its displacement)*sen(θ)*g and the final potential gravitational energy of the 30kg mass is 0.
the length of the string does not matter.

 

[RoboNerd]

forget about it, the solutions are alright.
what they do is they set an "imagined" length of string you don't need to know it, they just set the initial height of the 20kg mass to be the 0 of the referential, (and make it so its actual height is 20cm), so that the final potential energy of the 20kg mass is 0.2(its displacement)*sen(θ)*g and the final potential gravitational energy of the 30kg mass is 0.
the length of the string does not matter.

So it is a perfectly valid move to assume this... and how would I determine in a future problem that the length of the string does not matter, for future reference?

 

[WrongMan]

So it is a perfectly valid move to assume this... and how would I determine in a future problem that the length of the string does not matter, for future reference?

you have to conceptualize the problem...
in this case, imagine the triangle (slope) is on the floor, and the spring is underground, and its rest (unstreached) position makes it so the 30kg block is resting on the floor...
and you pull back on the 20kg block. you can make it so the original postion of the 20kg block minus its displacement makes it touch the floor, you have to understand that the postion of the 20kg block on the slope does not matter for the problem.
its just something you have to do, look at the problem form various angles until you find the best way to solve it...

and fyi the length of the string usually never matters in this sort of problems

 

[RoboNerd]

you have to conceptualize the problem...
in this case, imagine the triangle (slope) is on the floor, and the spring is underground, and its rest (unstreached) position makes it so the 30kg block is resting on the floor...
and you pull back on the 20kg block. you can make it so the original postion of the 20kg block minus its displacement makes it touch the floor, you have to understand that the postion of the 20kg block on the slope does not matter for the problem.
its just something you have to do, look at the problem form various angles until you find the best way to solve it...

and fyi the length of the string usually never matters in this sort of problems

Thanks so much for that idea!

I think I finally understand enough to do what the solutions says without feeling guilty about not understanding it!

 

[WrongMan]

Thanks so much for that idea!

I think I finally understand enough to do what the solutions says without feeling guilty about not understanding it!

much of solving physics problems is doing just that.
thinking about simpler systems that behave in the same way as the one beeing analysed

 

[faradayscat]

I just did the problem, you don't even need the height. Let "h" be the initial height of the 20-kg block (m). It's final height will be given by (h + 0.2). The initial height of the 30-kg block (M) is 0.4m (H) and its final height will be 0.2m (H*). Both the initial speeds are zero, the initial displacement of the string is 0.2m (x) and its final displacement is zero. Let their final speeds be "v". By conservation of energy:

mgh + MgH + ½kx² = mg(h+0.2) + MgH* + ½(m+M)v²

Do you see how the $mgh$ terms cancel? You don't need the height. Just pick the height as being "h" and it will cancel out anyway, or use any arbitrary value but that just complicates the calculation. Hope this helps!

 

[RoboNerd]

I just did the problem, you don't even need the height. Let "h" be the initial height of the 20-kg block (m). It's final height will be given by (h + 0.2). The initial height of the 30-kg block (M) is 0.4m (H) and its final height will be 0.2m (H*). Both the initial speeds are zero, the initial displacement of the string is 0.2m (x) and its final displacement is zero. Let their final speeds be "v". By conservation of energy:

mgh + MgH + ½kx² = mg(h+0.2) + MgH* + ½(m+M)v²

Do you see how the $mgh$ terms cancel? You don't need the height. Just pick the height as being "h" and it will cancel out anyway, or use any arbitrary value but that just complicates the calculation. Hope this helps!

Wow! Thanks so much! This will definitely help me in doing this homework!

 

[WrongMan]

I just did the problem, you don't even need the height. Let "h" be the initial height of the 20-kg block (m). It's final height will be given by (h + 0.2). The initial height of the 30-kg block (M) is 0.4m (H) and its final height will be 0.2m (H*). Both the initial speeds are zero, the initial displacement of the string is 0.2m (x) and its final displacement is zero. Let their final speeds be "v". By conservation of energy:

mgh + MgH + ½kx² = mg(h+0.2) + MgH* + ½(m+M)v²

Do you see how the $mgh$ terms cancel? You don't need the height. Just pick the height as being "h" and it will cancel out anyway, or use any arbitrary value but that just complicates the calculation. Hope this helps!

Also much of solving physics problems is just that.
inputing values as letters and pray to god they cancel :D

 

[faradayscat]

Also much of solving physics problems is just that.
inputing values as letters and pray to god they cancel :D

Well I have to agree with you on that one!