Acceleration of 1.44m/s^2: Solving Tension Problems

  • Thread starter Garrett21
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In summary, it seems that the pulley has to have specified mass and radius in order for the acceleration to be 1.44m/s^2.
  • #1
Garrett21
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Can someone please tell me how my teacher got Acceleration of 1.44m/s^2?
I don't see how its even possible without another force acting on the object
 

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  • #2
Garrett21 said:
Can someone please tell me how my teacher got Acceleration of 1.44m/s^2?
I don't see how its even possible without another force acting on the object

Is it possible that the pulley has specified mass and radius in the problem statement?

Chet
 
  • #3
Chestermiller said:
Is it possible that the pulley has specified mass and radius in the problem statement?

Chet

No, she is just introducing this so there is no friction.. I was scratching my head while everyone was saying they got the same answer following instructions, but i vaguely disagreed with it so that's why I am asking here.
 
  • #4
This seems to be a poorly designed problem. If you use F=ma and assume the pulley to be frictionless and massless you get a=(9.8)(5)/(15)=3.27 m/s^2

It seems that there is some assumed mass in the pulley though because they give you the distance it falls and the time it takes. Using those in the following kinematic equation yields an acceleration of 1.39 m/s^2 which is close:

x=x0+v0t+[itex]\frac{1}{2}[/itex]at2
 
  • #5
Garrett21 said:
No, she is just introducing this so there is no friction.. I was scratching my head while everyone was saying they got the same answer following instructions, but i vaguely disagreed with it so that's why I am asking here.
What were the instructions? I got the same results as oddjobmj.

Chet
 
  • #6
Chestermiller said:
What were the instructions? I got the same results as oddjobmj.

Chet

Just to find the acceleration at 1m... idk why this problem is annoying me so much but realistically on a test, i should write.. "this problem is not frictionless"
 
  • #7
One more thing. Are you sure the mass of A wasn't 30 Kg?

Chet
 
  • #8
Chestermiller said:
One more thing. Are you sure the mass of A wasn't 30 Kg?

Chet

Yes 100% if the mass was 30kg would the acceleration be equal to what the problem states?
 
  • #9
Garrett21 said:
Yes 100% if the mass was 30kg would the acceleration be equal to what the problem states?
Yes, pretty close.

chet
 

FAQ: Acceleration of 1.44m/s^2: Solving Tension Problems

1. What is acceleration?

Acceleration is the rate of change of an object's velocity over time. It can be calculated by dividing the change in velocity by the change in time.

2. How is acceleration measured?

Acceleration is typically measured in meters per second squared (m/s^2). This unit represents the change in velocity over one second.

3. How do you solve tension problems involving acceleration?

To solve tension problems involving acceleration, you must first identify the forces acting on the object and their directions. Then, you can use Newton's Second Law of Motion (F=ma) to calculate the tension force in the system.

4. What is the formula for acceleration?

The formula for acceleration is a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

5. How does the acceleration of 1.44m/s^2 affect an object's motion?

An acceleration of 1.44m/s^2 means that for every second an object is in motion, its velocity will increase by 1.44m/s. This will cause the object to speed up and cover more distance in less time.

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