Solving Dynamics Question: F_spring & F_spring Potential

  • Thread starter brk51
  • Start date
  • Tags
    Dynamics
In summary, the conversation discusses a problem involving a spring with a force constant of 500 N/m and a length of 0.5m. At an angle of 0 degrees, the spring force is 100 N. The individual is attempting to solve for the potential energy of the spring using the equation F_spring potential = 1/2ks^2, but is unsure why the force of the spring (100 N) is being plugged in instead of the potential energy. It is explained that the potential energy can be found by using the known force and length of the spring, resulting in a normal length of 0.3m and a stretched length of 0.2m.
  • #1
brk51
29
1

Homework Statement


Screen Shot 2017-06-18 at 1.34.48 PM.png


Homework Equations



F_spring = ks and F_spring potential = 1/2ks^2

The Attempt at a Solution



So I recognize that @ theta = 0 it is a 3/4/5 triangle. The length of the spring is at .5m...furthermore the given spring force at that angle is 100 N. So why is it wrong to do 100 = (1/2) (500) (s)^2 ==>> s = .632m... Why should I use F= ks

Everything else I understand...
 
Physics news on Phys.org
  • #2
Maybe I don't understand what you're trying to do, but it looks like you're plugging in the force of the spring where the potential energy should go.

If you know that the spring applies 500 Newtons for every meter its stretched, if it's applying 100 Newtons, it must have been stretched 0.2 meters (and therefore its normal length is 0.3 meters). That would allow you to find the potential energy in the spring. (Again, if I'm misunderstanding the problem, please tell me).
 

FAQ: Solving Dynamics Question: F_spring & F_spring Potential

What is the formula for calculating the force of a spring?

The formula for calculating the force of a spring is F = -kx, where F is the force, k is the spring constant, and x is the displacement from the equilibrium position.

How do I calculate the spring potential energy?

The formula for calculating the spring potential energy is PE = 1/2kx^2, where PE is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.

How does the force of a spring change with displacement?

The force of a spring is directly proportional to the displacement from the equilibrium position. As the displacement increases, the force also increases.

Can the force of a spring be negative?

Yes, the force of a spring can be negative. This indicates that the force is acting in the opposite direction of the displacement.

What is the relationship between spring force and spring potential energy?

The force of a spring and the potential energy are directly related. As the force increases, so does the potential energy. This is because a greater force is required to compress or stretch the spring, resulting in a higher potential energy.

Similar threads

Work / Energy Elevator / Spring Problem
Replies
5
Views
4K
Solving Disk Shaft (ENES102) Homework Equations
Replies
1
Views
1K
Solve Box-Spring Problem: Setup Differential Equation
Replies
16
Views
3K
Dynamics: Spring problem (Oscillation)
Replies
2
Views
2K
Inclined plane with pulley and spring
Replies
10
Views
2K
Can someone explain this derivation?
Replies
4
Views
2K
Spring-mass system with a pendulum using Lagrangian dynamics
Replies
15
Views
2K
Identifying the Potential Energy of a Spring Pulley System
Replies
1
Views
5K
Dynamics: Spring Question Solution
Replies
5
Views
6K
Dynamics pythagorean confusion
Replies
2
Views
2K

Hot Threads

Recent Insights

Back
Top