Length a Block on Pulley Moves - (Basic Energy Question)

[Alyssa Eiger]

Homework Statement

Two masses are connected by a string that is passed over a massless pulley. At t=0, the 6-kg mass is moving upwards at 2m/s, while the 4-kg mass is descending at the same speed.

How far will the 6-kg mass rise before it stops?

Homework Equations

KE= 1/2*m*v^2
PEg: m*g*h

The Attempt at a Solution

- I considered the entire system (both blocks) as the system, with no external forces acting - only gravity is at play.
- Same speed is mentioned, so there is no acceleration.
- The tension between the two blocks is the same, as the string is massless.
- I determined that initially: there is KE and PEg - and final state only has PEg. I set both equal to each other, (as indicated on the attached photo tweet), but am having a difficult time determining next-steps.

Any input is appreciated.

 

[Alyssa Eiger]

https://twitter.com/wrongedauthor/status/1001147259503890432

 

[Orodruin]

Same speed is mentioned, so there is no acceleration.

This is not correct. The blocks having the same speed does not mean that they cannot change speed.

I determined that initially: there is KE and PEg - and final state only has PEg. I set both equal to each other, (as indicated on the attached photo tweet), but am having a difficult time determining next-steps.

Please post your work in text.

 

[scottdave]

They will be traveling at the same rate, for any given time, but there will be acceleration. I would draw a Free Body Diagram for each block, then do as you said by setting the magnitude of the tensions equal (and the magnitudes of the velocities equal).

 

[scottdave]

You could call the location of the 6 kg block (at time 0) to be location zero, so the potential energy of that block is zero there. You could also set the smaller block up to have zero PE at a certain point to maybe help your calculations.

 

[Alyssa Eiger]

@scottdave @Orodruin - thank you for your invaluable input.

Initial States:
6KG Block: If I set the 6-kg block at t=0 to be location zero, with a potential energy of zero - then all my energy at that point in time for the block is KE (1/2*6kg*4m/s) = 12 Joules.

4KG Block: At the initial point t=0, there is both KE and PEg. KE is: (1/2)(4kg)(4m/s) = 8 Joules. There is also PEg: (4kg)(10m/s)(h) which is equal to 4 Joules (the difference between 12J-8J). If PEg = 4Joules, then: 4J = (4kg)(10m/s)(h). Height = .1m

Final States:
6KG Block: When the block comes to rest - the only energy acting is PEg - also at 12 J.
4KG Block: When the block comes to rest - the only energy acting is PEg - also at 12 J.Photo of my FBD: https://twitter.com/wrongedauthor/status/1001152490635644929

I also don't quite understand how the FBD is possible, if the 6KG block is moving UP - the force of the tension on the strong has to be greater than 60N - but that is the same value of tension on the 4KG block, which is larger than the Weight of the Earth on the block (40N) - which indicates there's no way the 4kg block could move down.

The back of the book reports the answers is 1M, but all my calculations have either reflected .2M or .1M. This is my first physics course.

 

[Orodruin]

6KG Block: When the block comes to rest - the only energy acting is PEg - also at 12 J.
4KG Block: When the block comes to rest - the only energy acting is PEg - also at 12 J.

There is no reason to assume that each block has the same total energy as it had before. The blocks by themselves do not form a closed systen.

I also don't quite understand how the FBD is possible, if the 6KG block is moving UP - the force of the tension on the strong has to be greater than 60N

This is not correct. The tension can be lower than 60 N while it moves up. All this tells you is that the acceleration is in the opposite direction compared to the velocity - which it must be to eventually stop!

Anyway, energy considerations should be sufficient. There is no reason to start considering the internal forces and drawing FBDs.

 

[scottdave]

...Anyway, energy considerations should be sufficient. There is no reason to start considering the internal forces and drawing FBDs.

Yes I agree, after rereading the problem.