Solving Spring Constant and Elastic Energy Problems

In summary, the conversation discusses a mass of 100g suspended from a light spring and its extension of 4cm when in equilibrium. It also covers finding the spring constant, calculating elastic energy, determining the extension of a spring with a different mass, and the relationship between work and extending a spring. The equations used include F=ma, k=F/Δx, and Ep=1/2FΔx.
  • #1
lemon
200
0
1. A mass of 100g is suspended from a light spring and extends the spring by 4cm when the mass in in equilibrium.
a) Find the spring constant k of the spring?
b) How much elastic energy is now stored in the spring?
c) What would be the extension of the spring if a mass of 170g were suspended?
d) How much extra energy would now be stored in the spring?
e) Why is more work needed to stretch a spring by 1cm from its unextended length, than is required to stretch it by an extra 1cm when the spring is already extended?



2. Let deltaX = e
F=ke
Ep=1/2Fe
Stress=F/A
Strain=e/L
E=stress/strain
F=ma

100g = 0.1kg
4cm = 0.04m
Let gravity = -10m/s



3. a) F=0.1 x 10 = 1N
k=10N/0.04m = 250N/m

b) E=1/2 x 10N x 0.04m = 0.2J
c) F=0.170 x 10 = 1.7N
e = 1.7N/250N/m = 0.006.8m
d) E = 1/2 x 10N x 0.170m = 0.85J
e) I know this holds true from my experience in reality, but when I consider the force extension graph and hook's law, it tells me that up until the elastic limit the same force is required to achieve the same extension. But considering E=1/2 x stretching force x extension, from this equation it is clear that an increase in extension will increase elastic potential energy.


Could someone kindly check and guide, please?
Thank you :bugeye:
 
Physics news on Phys.org
  • #2
Is there something wrong with this thread. Many views but no reply. Is it cause it's such easy question?
 
  • #3
Ok. You'll better ignore the above post. I guess I really was under too much stress and strain.
So, hopefully to put it right:

mass=0.10kg
Δx=0.04m

a) Find the spring constant k of the spring?
F=ma
0.10x10=1.0N

F=kΔx
k=F/Δx
1.0/0.04=25N/m
 
  • #4
b) How much elastic energy is now stored in the spring?
Ep=1/2FΔx
1/21.0x0.04=0.02J

c) What would be the extension of the spring if a mass of 170g were suspended?
F=ma
0.170kgx10=1.70N
Δx=F/k
1.70/25=0.068m

d) How much extra energy would now be stored in the spring?
Ep=1/2FΔx
1/2x1.70x0.68=0.578m
 

What is a spring constant?

A spring constant, also known as a force constant, is a measure of the stiffness or resistance of a spring to being compressed or extended. It is represented by the letter k and is measured in units of force per unit length.

How is the spring constant calculated?

The spring constant can be calculated by dividing the force applied to the spring by the displacement of the spring from its resting position. This is known as Hooke's Law, which states that the force applied to a spring is directly proportional to its displacement.

What is elastic potential energy?

Elastic potential energy is the energy stored in an object when it is deformed or stretched. In the case of a spring, this energy is stored in the form of potential energy due to the displacement of its coils from their resting position.

How is elastic potential energy calculated?

The elastic potential energy of a spring can be calculated using the formula E = 1/2 * k * x^2, where E is the elastic potential energy, k is the spring constant, and x is the displacement of the spring from its resting position.

What are some real-life applications of spring constant and elastic energy?

Spring constants and elastic energy are important in many industries and technologies, including engineering, construction, and transportation. They are used in the design of springs for various mechanical devices, such as car suspension systems and mattresses. They are also crucial in the development of renewable energy sources, such as wind turbines, which use springs to store and release elastic energy.

Similar threads

  • Introductory Physics Homework Help
Replies
17
Views
213
  • Introductory Physics Homework Help
Replies
29
Views
785
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
290
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
767
  • Introductory Physics Homework Help
Replies
2
Views
949
  • Introductory Physics Homework Help
Replies
20
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
815
Back
Top