Static equilibruim in light spring

In summary, a 0.20 kg light spring is suspended from its lower end and a second mass of 0.10 kg is suspended from the first by a thread. When the arrangement is burned through, the upward acceleration of the 0.20 kg mass is 5.0 m s^-2, considering g=10 m s^-2. The total weight acting on the first spring is 3 N and the force and acceleration of the upper mass can be calculated by taking out the tension of the cut string.
  • #1
jinx007
62
0
A light spring has a mass of 0.20 kg suspended from its lower end. A second mass of 0.10 kg is suspended from the first by a thread. The arrangement is allowed to come into static equilibrium and then is burned through. At this instant, what is the upward acceleration of the 0.20 kg mass. ( take g= 10 m s^-2)

Answer according to the booklet is 5.0 m s^-2

total weight acting on the first spring is 3 N ( i think so..)

Then i don't really know how to proceed...help..!
 
Physics news on Phys.org
  • #2


You can do this : look at the force on the upper mass just before the string is burnt through since the spring continues to exert the same force as long as the extension does not change (which does not at that instant as the problem says), all you have to do is take out the tension of the now cut string and you will have the force and acc.
 

What is static equilibrium in light spring?

Static equilibrium in light spring refers to a state where the forces acting on a spring are balanced, resulting in no net force or acceleration. This means that the spring is at rest or moving at a constant velocity.

What factors affect static equilibrium in light spring?

The factors that affect static equilibrium in light spring include the mass of the object attached to the spring, the stiffness of the spring, and the gravitational force acting on the object.

How do you calculate the force in a light spring at static equilibrium?

The force in a light spring at static equilibrium can be calculated using the equation F = kx, where F is the force in Newtons, k is the spring constant, and x is the displacement of the spring from its equilibrium point.

What is the significance of static equilibrium in light spring?

Static equilibrium in light spring is important because it allows us to understand the behavior of objects attached to springs. It is also used in various applications such as measuring forces, designing structures, and understanding the mechanics of machines.

How does the concept of static equilibrium in light spring apply to real-life situations?

The concept of static equilibrium in light spring applies to real-life situations such as weighing scales, trampolines, and shock absorbers. It also helps in understanding the stability of structures like bridges and buildings.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
863
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
16
Views
812
  • Introductory Physics Homework Help
Replies
26
Views
2K
Replies
6
Views
779
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
905
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
2K
Back
Top