Maximum Acceleration with Given Mass and Tension?

[kamalpreet122]

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 :-(

 

[Apphysicist]

Don't forget gravity!

 

[doublemint]

You have to accout for the acceleration due to gravity!

Edit: Your too fast for me Apphysicist !

 

[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.

 

[kamalpreet122]

so the formula would be changed to F= (mg)a ?

 

[Apphysicist]

so the formula would be changed to F= (mg)a ?

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

Do you see why?

 

[kamalpreet122]

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

 

[kamalpreet122]

man yur great =P i have a really tough question i was not able to figure out want to help me with that :$

 

[Apphysicist]

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:

\sum_{\text{all}}\bold{F}\,=\,m \bold{a}<br />

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).

 

[kamalpreet122]

ohhh Thnxxx :D