Find the acceleration of the system (2 blocks sliding on a table)

In summary: Ft = 50NCase 2- FBD od 5 kg...Ft = 50NIn both the cases, the tension force is the same. However, in case 1, the surface is taken as frictionless, so the net force is greater.
  • #1
paulimerci
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Homework Statement
A 7kg block rests initially on a table in case 1 and case 2 above. The coefficient of friction between the block and the table is 0.2. The block is attached by a cord of negligible mass which hangs over a frictionless pulley. In case 1 a force of 50N is exerted on the pulley and in case 2, a 5kg mass is hung. Calculate the acceleration of the system in case 1 and in case 2.
Relevant Equations
F = ma
In both the cases 7 kg mass accelerates towards the right because of the 50N force. The unbalanced forces in both the cases are the force of gravity due to 5kg block and force of friction. Applying Newton's second law of motion to cases 1 and 2 yields the following results for acceleration:
Where m1=5kg, m2 = 7kg, g=10m/s2
Case 1,
a = Net force/ total mass

a = m1g - friction force/(m1+m2)
= m1g - friction force/ (m1+m2)
= 50 - (0.2x7x10)/12 = 3m/s2

Case2,
a = Net force/ total mass

a = m1g - friction force/(m1+m2)
= 50 - (0.2x7x10)/12 = 3m/s2

In both the cases acceleration came out to be the same. But the textbook gives a different answer where Case 1 acceleration is greater than Case 2. Does that mean in Case 1 the surface is taken as frictionless?
 

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  • #2
paulimerci said:
Homework Statement:: A 7kg block rests initially on a table in case 1 and case 2 above. The coefficient of friction between the block and the table is 0.2. The block is attached by a cord of negligible mass which hangs over a frictionless pulley. In case 1 a force of 50N is exerted on the pulley and in case 2, a 5kg mass is hung. Calculate the acceleration of the system in case 1 and in case 2.
Relevant Equations:: F = ma

In both the cases 7 kg mass accelerates towards the right because of the 50N force. The unbalanced forces in both the cases are the force of gravity due to 5kg block and force of friction. Applying Newton's second law of motion to cases 1 and 2 yields the following results for acceleration:
Where m1=5kg, m2 = 7kg, g=10m/s2
Case 1,
a = Net force/ total mass

a = m1g - friction force/(m1+m2)
= m1g - friction force/ (m1+m2)
= 50 - (0.2x7x10)/12 = 3m/s2

Case2,
a = Net force/ total mass

a = m1g - friction force/(m1+m2)
= 50 - (0.2x7x10)/12 = 3m/s2

In both the cases acceleration came out to be the same. But the textbook gives a different answer where Case 1 acceleration is greater than Case 2. Does that mean in Case 1 the surface is taken as frictionless?
What is the force of tension from the hanging weight when the system is accelerating at ##a##?

Present the following:

Case 1- FBD od 7 kg block.

Case 2 - FBD of 7 kg Block and 5 kg Block for case 2.
 
Last edited:
  • #3
paulimerci said:
Does that mean in Case 1 the surface is taken as frictionless?
No.

paulimerci said:
a = m1g - friction force/(m1+m2)
= m1g - friction force/ (m1+m2)
= 50 - (0.2x7x10)/12 = 3m/s2
Brackets, brackets, brackets !
and 3m/s2 looks horrible too ! ##3 ## m/s2 !

What is m2 doing in your equations for case 1 ? :nb)

##\ ##
 
  • #4
In case 1 what does m1g represent? There is only one mass and that the 7 kg mass on the table. It is pulled by a constant force of 50 N. What is Newton;s second law in this case?
 
  • #5
BvU said:
No.Brackets, brackets, brackets !
and 3m/s2 looks horrible too ! ##3 ## m/s2 !

What is m2 doing in your equations for case 1 ? :nb)

##\ ##
I see that things are a bit more confused than what my gut thought was error. :olduhh:
 
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  • #6
Lets work in all variables.

##M## = 7 kg mass

##m##= 5 kg mass

Start with case 1:

What is Newtons Second law for the mass ##M## in the ## x \rightarrow^+ ##

Please, take a minute and try to format the equations using latex. It is not difficult to learn.
 
  • #7
kuruman said:
In case 1 what does m1g represent? There is only one mass and that the 7 kg mass on the table. It is pulled by a constant force of 50 N. What is Newton;s second law in this case?
 

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  • #8
That's better. What about case 2?
 
  • #9
kuruman said:
That's better. What about case 2?
 

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  • #10
That's good. Congratulations, you've done it. Now please consider learning how to use LaTeX.
 
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  • #11
For future reference its always better to just state your directional conventions so we don't have to reverse engineer it from the equations you provide.
 
  • #12
kuruman said:
That's good. Congratulations, you've done it. Now please consider learning how to use LaTeX.
Thank you, Will try!
 
  • #13
erobz said:
For future reference its always better to just state your directional conventions so we don't have to reverse engineer it from the equations you provide.
okay! Thank you!
 
  • #14
paulimerci said:
Thank you, Will try!
I'm trying to add equations using LateX and I see the preview icon is not working and is there any way to check the equations that I entered in LateX mode is correct?
 
  • #15
paulimerci said:
I'm trying to add equations using LateX and I see the preview icon is not working and is there any way to check the equations that I entered in LateX mode is correct?
When this happens to me I reload the page.
 
  • #16
kuruman said:
When this happens to me I reload the page.
Thanks I did, still not working.
 
  • #17
paulimerci said:
I'm trying to add equations using LateX and I see the preview icon is not working and is there any way to check the equations that I entered in LateX mode is correct?
They won't load until someone post some Latex in the thread.

##F = m a ##

Now, preview should work for you.
 
  • #18
erobz said:
They won't load until someone post some Latex in the thread.

##F = m a ##

Now, preview should work for you.
Are there any ways to check online without posting them?
 
  • #19
paulimerci said:
Are there any ways to check online without posting them?
Is the preview still not working?
 
  • #20
erobz said:
Is the preview still not working?
It's not working.
 
  • #21
Did you use the delimiters # (doubled) and $ (doubled)? I can't think of anything else.
 
  • #22
kuruman said:
Did you use the delimiters # (doubled) and $ (doubled)? I can't think of anything else.
I tried $(doubled) first, but it didn't work. I then tried $(single), but that didn't either.
 
  • #23
paulimerci said:
Thanks I did, still not working.
To be clear, it can be necessary to reload the page and then toggle preview a couple of times. If it is still not working, please just post some anyway so that we can see if you are doing it right.
 
  • #24
haruspex said:
To be clear, it can be necessary to reload the page and then toggle preview a couple of times. If it is still not working, please just post some anyway so that we can see if you are doing it right.
As of right now, it's fixed. I'll let you know if that changes. Thank you.
 

FAQ: Find the acceleration of the system (2 blocks sliding on a table)

What is acceleration?

Acceleration is the rate of change of velocity over time. It is a vector quantity, meaning it has both magnitude and direction.

How do you find the acceleration of a system?

To find the acceleration of a system, you need to first calculate the net force acting on the system. Then, use Newton's second law of motion (F=ma) to determine the acceleration, where F is the net force and m is the mass of the system.

What is the difference between acceleration and velocity?

Velocity is the rate of change of an object's position over time, while acceleration is the rate of change of an object's velocity over time. In other words, velocity tells us how fast and in what direction an object is moving, while acceleration tells us how much an object's velocity is changing.

How does the mass of the objects affect the acceleration of the system?

The mass of the objects affects the acceleration of the system because it is a factor in Newton's second law of motion. The greater the mass, the greater the force needed to accelerate the object, resulting in a smaller acceleration. This means that the acceleration of a system with two blocks of different masses will be different, even if the net force acting on the system is the same.

Can the acceleration of a system be negative?

Yes, the acceleration of a system can be negative. This indicates that the system is decelerating or slowing down in the direction of motion. It can also mean that the system is accelerating in the opposite direction of its motion.

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