Potential energy in case of Atwood machine

[rudransh verma]

Homework Statement

The heavier block in an Atwood machine has a mass twice that of lighter one. The tension in the string is 16N when the system is set into motion. Find the decrease in U(GPE) during the first sec after the system is released from rest.

Relevant Equations

$F_{net}=ma$
$s=ut+1/2at^2$
$\Delta U=mgh$

$T-2mg=2ma_1$ (acceleration of heavier mass)
$T-mg=ma_2$
($-a_1=a_2$)

On solving the eqns, $a_1=-g/3=-a_2$

$s=1/2at^2$
$s=-g/6$ , distance covered by heavier mass.
$s=g/6$ , covered by lighter mass.

Edit: $\Delta U_1=mgh=-2mg^2/6$ (decrease in U of heavier mass)
$\Delta U_2=mgh=mg^2/6$ (increase in U of lighter mass)
$\Delta U=\Delta U_1+\Delta U_2=-mg^2/6$

Now what is m?

 

[Orodruin]

You have not used all of the information in the problem.

 

[Steve4Physics]

You haven't yet told us what you think it is ! (Or how you arrived at $\Delta U=-mg^2/6$)

 

[rudransh verma]

You haven't yet told us what you think it is ! (Or how you arrived at $\Delta U=-mg^2/6$)

I don’t know m. It’s not given. Also I have edited the post. I thought it’s pretty clear but anyways.

 

[haruspex]

I don’t know m.

Because

You have not used all of the information in the problem.

 

[Steve4Physics]

I don’t know m. It’s not given. Also I have edited the post. I thought it’s pretty clear but anyways.

The question should, ideally, state what $m$ is. But the author has assumed it is clear from the information given in the question.

So, as already suggested by @Orodruin and @haruspex, read the question again and take a guess at what you think $m$ is. What is your best guess?

(You can then find $\Delta U$ based on your guess.)

Sorry @rudranch verma. Based on your original (pre-editted) Post #1, I thought you were asking 'what does $m$ represent?

But it looks like you want the value of $m$. See @Orodruin's and @haruspex's posts.

 

[rudransh verma]

You have not used all of the information in the problem.

Got it! 4mg/3=16, m= 1.22 kg
$\Delta U= 19.6 J$
Thank you!
It’s not very clear when is T=16N but I think it’s during motion. That’s how I wrote the eqns.

 

[rudransh verma]

@rudranch verma. Based

It’s Rudransh.

 

[Steve4Physics]

It’s Rudransh.

In Post #6 I typed the @ symbol followed by your PF user name: rudranch verma.

The system should have automatically converted it into a link, and rendered it blue. It didn't!

I'll try it again here: @rudransh verma

Yea - seems to work now.

 

[Nugatory]

Also I have edited the post.

Do not edit posts after someone has replied. If you want to add information to your post or provide a necessary clarification, do so in a followup post.

Editing a post after it has become part of an active discussion is unfair to the people who are volunteering their help, makes it impossible for anyone else to follow the thread, and is fundamentally dishonest.

 

[rudransh verma]

Do not edit posts after someone has replied. If you want to add information to your post or provide a necessary clarification, do so in a followup post.

I always cut the sentences or mention edit: I didn’t do that in this thread. I forgot. But I’ll mention now.

In Post #6 I typed the @ symbol followed by your PF user name: rudranch verma.

Ranch is a sauce. Rudra(shiva)-nsh(part) means “Part of Shiva”

 

[Orodruin]

I always cut the sentences or mention edit: I didn’t do that in this thread. I forgot. But I’ll mention now.

As stated: Just don’t edit the first posts in threads that have answers.

 

[Steve4Physics]

Ranch is a sauce. Rudra(shiva)-nsh(part) means “Part of Shiva”

Apologies. I only just realized that I mistyped 'ch' instead of 'sh'.

 

[Lnewqban]

Got it! 4mg/3=16, m= 1.22 kg
$\Delta U= 19.6 J$
...

Why 4mg?

 

[Orodruin]

Why 4mg?

Because 1 + 1/3 = 4/3.

 

[Vanadium 50]

It is better to follow the rules than to explain why they don't apply to you.

This whole thread has been an exercise in guessing. Perhaps this problem is at too high a level?