Instantaneous acceleration of blocks (1 Viewer)

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1. The problem statement, all variables and given/known data

The blocks are of mass 2 kg shown is in equilibrium. At t = 0 right spring in fig (i) and right string in
fig (ii) breaks. Find the ratio of instantaneous acceleration of blocks?


2. Relevant equations



3. The attempt at a solution

In fig (i) before the string is cut,(2kx)sin37° = 2g , kx=g/sin37°

In fig (ii) before the string is cut,(2T)sin37° = 2g , T=g/sin37°

Is the instantaneous acceleration in the two cases same i.e ratio 1:1


I would be grateful if somebody could help me with the problem .
 

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BvU

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I know physics isn't democracy, but I'm inclined to vote for 1:1 too!
Net force after cutting will be minus the missing force, and these are equal.
I'll be listening in to see if someone can prove me (us) wrong!
 

PhysicoRaj

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What is the state of the spring at t=0? Stretched? Equilibrium? Compressed?
 
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I know physics isn't democracy, but I'm inclined to vote for 1:1 too!
Net force after cutting will be minus the missing force, and these are equal.
I'll be listening in to see if someone can prove me (us) wrong!
I don't think so. When the string is cut, the acceleration of the mass is purely tangential. When the spring is cut, the acceleration is not purely tangential.

Chet
 
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I don't think so. When the string is cut, the acceleration of the mass is purely tangential. When the spring is cut, the acceleration is not purely tangential.

Chet
Excellent thinking!!!

Are you getting 25/24 ?
 
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Thanks Chet :smile:
 

BvU

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Impressed ! :redface:
 
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Tanya, can you please show how you solved this problem?

Thanks!
In the first case i.e having springs

From equilibrium condition kx = 5g/3

After string is cut , the acceleration of the block will have both radial and tangential components .

at = 2gcos37° = 8g/5

ar = kx - 2gsin37° = 5g/3 - 6g/5 = 7g/15

∴ net acceleration a1 = 5g/3

In the second case i.e having strings

From equilibrium condition T = 5g/3

After string is cut , the acceleration of the block will be purely tangential i.e ar = 0

at = 2gcos37° = 8g/5

∴ net acceleration a2 = 8g/5


So a1/a2 = 25/24
 
3,811
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In the first case i.e having springs

From equilibrium condition kx = 5g/3

After string is cut , the acceleration of the block will have both radial and tangential components .

at = 2gcos37° = 8g/5

ar = kx - 2gsin37° = 5g/3 - 6g/5 = 7g/15

∴ net acceleration a1 = 5g/3

In the second case i.e having strings

From equilibrium condition T = 5g/3

After string is cut , the acceleration of the block will be purely tangential i.e ar = 0

at = 2gcos37° = 8g/5

∴ net acceleration a2 = 8g/5


So a1/a2 = 25/24
Excellent, thanks a lot Tanya! :smile:
 

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