Calculate Mass Acceleration with Spring Length Change of 0.02 m

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a spring and a mass. Initially, a mass of 77.4 kg stretches a spring by 0.14 meters when hung vertically. The problem then transitions to a scenario where the same mass is placed on a horizontal surface, and the spring is stretched by 0.02 meters. Participants are tasked with determining the acceleration of the mass in this new context.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the forces acting on the mass and the implications of the spring's length change. There are inquiries about how to apply Hooke's Law and the relationship between the two different spring stretches.

Discussion Status

Some participants have offered guidance on using the initial conditions to find the spring constant and applying Hooke's Law to determine the force acting on the mass. There is an ongoing exploration of the differences between the two scenarios and how they affect the calculations.

Contextual Notes

There is a noted confusion regarding the application of the spring stretches from two different situations, with participants questioning the relevance of the initial stretch when calculating the acceleration in the second scenario.

buffgilville
Messages
91
Reaction score
0
A spring hanging vertically is stretched by 0.14 Meters when a mass of 77.4 kg is hung from it. The mass is then placed on a smooth horizontal surface and pulled using the same spring. If the spring is now stretched by 0.02 Meters, what is the acceleration of the mass (in M/s2).

I used the formula F=ma and got 9.81. I got it wrong. What am suppose to do with the spring length change?
 
Physics news on Phys.org
In the first case what forces are acting on the block?
 
buffgilville said:
What am suppose to do with the spring length change?

Remember Hooke's Law.
 
F or T on the box: T=mg = 77.4 * 9.81 = 759.294
Hookes law: F = -kx so, 759.294/(0.14-0.02) = -6327.45
how do I find the acceleration of the mass from here?
 
buffgilville, do you know what the x in Hooke's Law mean?
 
x = direction moved
 
A spring hanging vertically is stretched by 0.14 Meters when a mass of 77.4 kg is hung from it. The mass is then placed on a smooth horizontal surface and pulled using the same spring. If the spring is now stretched by 0.02 Meters, what is the acceleration of the mass (in M/s2).

Do you see a problem with what you did?
 
I know I did the problem wrong, but I don't know what? I don't know how to find the acceleration for this type of problem. Can you please explain it to me?
 
Write what you need, and then read all i said, and see if you can understand. If not try to read Hooke's Law again, and see what [itex]\vec{F} = k \vec{x}[/itex] means.
 
  • #10
buffgilville said:
I know I did the problem wrong, but I don't know what? I don't know how to find the acceleration for this type of problem. Can you please explain it to me?

Maybe this will help - in your solution, you subtracted the .02 m from the .14 m. The .14 m came from when the mass was hung from the spring. The .02 m came from when you dragged the mass along a surface. Those were two completely different situations - why would the two distances have anything to do with each other?

On the other hand, there were some things about the two situations that remained exactly the same between them - those are the things that you'll use from the first problem to solve the second. I realize that may be a little cryptic, but ask yourself - what did not change from the first part to the second?
 
  • #11
buffgilville said:
I know I did the problem wrong, but I don't know what? I don't know how to find the acceleration for this type of problem. Can you please explain it to me?
First use the information provided when the mass is hanging vertically to find the spring constant. Then use Hooke's law to find the force on the mass when it's on the horizontal surface. Finally, use Newton's 2nd law to find the acceleration of the mass.
 
  • #12
Thanks a million for all of your help!
 

Similar threads

Replies
17
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
14
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
16
Views
2K
Replies
1
Views
3K