Calculate Mass Acceleration with Spring Length Change of 0.02 m

In summary, the mass is accelerated by 6327.45 kg/s when it is pulled using the same spring that was stretched by .14 meters when the mass was hung from it.
  • #1
buffgilville
91
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
  • #2
In the first case what forces are acting on the block?
 
  • #3
buffgilville said:
What am suppose to do with the spring length change?

Remember Hooke's Law.
 
  • #4
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?
 
  • #5
buffgilville, do you know what the x in Hooke's Law mean?
 
  • #6
x = direction moved
 
  • #7
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?
 
  • #8
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?
 
  • #9
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!
 

1. What is the formula for calculating mass acceleration with a spring length change of 0.02 m?

The formula for calculating mass acceleration with a spring length change of 0.02 m is: a = (k/m) * x, where a is the acceleration in meters per second squared (m/s^2), k is the spring constant in Newtons per meter (N/m), m is the mass in kilograms (kg), and x is the change in spring length in meters (m).

2. How do you determine the spring constant (k) for this calculation?

The spring constant (k) can be determined by dividing the force applied to the spring (in Newtons) by the change in spring length (in meters). This will give you the spring constant in N/m, which is needed for the calculation.

3. Can this calculation be applied to any spring, regardless of its size or material?

Yes, this calculation can be applied to any spring as long as the spring constant (k) and the mass (m) are known. The spring constant may vary depending on the size and material of the spring, but the formula remains the same.

4. Is there a specific unit for the result of this calculation?

The result of this calculation will be in meters per second squared (m/s^2), which is the unit for acceleration. This unit measures the change in velocity per unit of time.

5. How can this calculation be used in real-world applications?

This calculation can be used in various real-world applications, such as determining the acceleration of an object attached to a spring, measuring the force applied to a spring, or designing and testing springs for use in different mechanisms or devices.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
339
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
16
Views
812
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
987
  • Introductory Physics Homework Help
Replies
8
Views
3K
Back
Top