Grade 11 Physics help

  • #1
Suppose a spring is attached vertically to the ceiling of an elevator. When a 5.00Kg mass is attached to the spring, it descends a distance of 25.5 cm.

a) what is the spring constant of the spring?

I found out that it was 192 N/m, so my problem is with part B!

b) how far would the spring stretch if the elevator were accelerating up at a rate of 3.50 m/s2?

I just want to know what I should do, because I don't know! Thanks!
 

Answers and Replies

  • #2
so the work I did in order to determine the spring constant is as follows:

F=KX

K=F/X=mg/x=(5.00kg)(9.81m/s2)/0.255m=192N/m

I don't understand part b though!
 
  • #3
114
0
The spring is now counteracting gravity as well as the acceleration of the elevator away from the mass.So the total force is

[itex]
F_{total} = m(g + a_{elevator})
[/itex]
set this equal to the equation of how the spring reacts to a force
[itex]
F_{total} = KX
[/itex]

and you'll have your distance
 
  • #4
thanks, but for the Ftotal, wouldn't it be equal to m(g-aelevator) since it is going opposed to it?
 
  • #5
114
0
np man, happy to help!

since the elevator is going up and away from the mass it will additionally stretch the spring

It's one of those things that you have to work your head around. If you are in an elevator and it goes up you are heavier as it accelerates.

As you go down in an elevator you are lighter (you feel your stomach go up a little).

So if the elevator was going down the acceleration would be negative.
 
  • #6
alright, I understand! Thanks for the help!
 

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