Maximum Acceleration with Given Mass and Tension?

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Homework Help Overview

The problem involves a mass being pulled upwards by a cable with a specified maximum tension, and participants are discussing how to determine the maximum acceleration while considering the effects of gravity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law (F=ma) and the need to account for gravitational force when calculating net force and acceleration. There are questions about the correct formulation of the forces involved.

Discussion Status

Participants are actively engaging with the problem, clarifying the role of gravity in the calculations and exploring how to set up the equations correctly. Some guidance has been provided regarding the need to consider net forces.

Contextual Notes

There is an emphasis on ensuring that gravitational effects are included in the calculations, and participants are questioning their assumptions about the forces acting on the mass.

kamalpreet122
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Homework Statement



A 2.00X101 kg mass is pulled upwards with constant acceleration by a cable attached to a motor. The cable can withstand a maximum tension of 5.00X102 N. What is the maximum acceleration possible?

The answer sheet says the answer is 15.2m/s2

Homework Equations



F=ma

The Attempt at a Solution



F=ma
F/m = a
500/20 = a
25 m/s2= a

some1 please tell me what i did wrong :-(
 
Physics news on Phys.org
Don't forget gravity!
 
You have to accout for the acceleration due to gravity!

Edit: Your too fast for me Apphysicist !
 
Indeed! F=ma is your net force = m* net acceleration. If you apply your maximum tension upward, gravity still acts downward (sum the forces for your net force, don't forget that gravity gets a minus-sign because it acts in the opposite direction of your upward tension). Then use F=ma.
 
so the formula would be changed to F= (mg)a ?
 
kamalpreet122 said:
so the formula would be changed to F= (mg)a ?

No, it would be changed to (T-mg)=ma.

Do you see why?
 
yeaaa kinda ..soo the total weight subtracted from Tension gives me the net force then i apply the forumlaa .. OH that makes so much senseee =) THNXXX :D
 
man yur great =P i have a really tough question i was not able to figure out want to help me with that :$
 
kamalpreet122 said:
yeaaa kinda ..soo the total weight subtracted from Tension gives me the net force then i apply the forumlaa .. OH that makes so much senseee =) THNXXX :D

Right. It's much easier to think about F=ma really being:

[tex]\sum_{\text{all}}\bold{F}\,=\,m \bold{a}[/tex]

So it's a sum of all your vector forces (the bold meaning it's a vector...so for your problem, the +/- indicates the direction is all, since there are no components, it's just in one direction).
 
  • #10
ohhh Thnxxx :D
 

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