Elastic Potential Energy Stored

In summary, the elastic potential energy stored in the stretched band with a mass of 0.67g, initial horizontal velocity and initial position of 9.6 cm above the floor, and a horizontal displacement of 3.6 m, can be calculated by using the Law of Conservation of Energy and relating the velocity to the elastic energy through a projectile problem.
  • #1
kevykevy
25
0
Question - A stretched elastic band of mass 0.67g is released so that its initial velocuty is horizontal, and its initial position is 9.6 com above the floor. What was the elastic potential energy stored in the stretched band if, it has a horizontal displacement of 3.6 m from its initial position?

Solution

Find v
At finish (h = 0)
Et = Ek + Eg
Et = 0.000335v^2

At Start (h = 0.096m)
Et = Ek + Eg
Et = 0.000335v^2 + 0.0006309792
by Law of Conservation of Energy (LCE)
Et = 0.000335v^2
0.000335v^2 = 0.000335v^2 + 0.0006309792

so when I got there, I was stuck and thought I went down the wrong track

But my method was going to be to find the velocity so I could find Ek which equals Ee because:

W = Ek
W = 1/2 k x^2
Ee = 1/2 x^2
so
Ee = Ek
 
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  • #2
kevykevy said:
Question - A stretched elastic band of mass 0.67g is released so that its initial velocuty is horizontal, and its initial position is 9.6 com above the floor. What was the elastic potential energy stored in the stretched band if, it has a horizontal displacement of 3.6 m from its initial position?

Solution

Find v
At finish (h = 0)
Et = Ek + Eg
Et = 0.000335v^2

At Start (h = 0.096m)
Et = Ek + Eg
Et = 0.000335v^2 + 0.0006309792
by Law of Conservation of Energy (LCE)
Et = 0.000335v^2
0.000335v^2 = 0.000335v^2 + 0.0006309792

so when I got there, I was stuck and thought I went down the wrong track

But my method was going to be to find the velocity so I could find Ek which equals Ee because:

W = Ek
W = 1/2 k x^2
Ee = 1/2 x^2
so
Ee = Ek
Your last appraoch will relate the velocity the band acquires at the intitial height to the elastic energy. Then it is a projectile problem to find that velocity.
 
  • #3
= 1/2 mv^2

The elastic potential energy stored in a stretched elastic band can be calculated using the formula Ee = 1/2 kx^2, where k is the spring constant and x is the displacement from the equilibrium position. In this case, the mass of the elastic band is 0.67g and its initial displacement is 9.6 cm (0.096 m) above the floor. We are given that the elastic band has a horizontal displacement of 3.6 m from its initial position, and we need to find the elastic potential energy stored at this point.

To do this, we first need to find the spring constant of the elastic band. This can be done by using the formula k = F/x, where F is the force applied and x is the displacement. In this case, the force applied is the weight of the elastic band, which can be calculated using the formula F = mg, where m is the mass of the elastic band and g is the acceleration due to gravity (9.8 m/s^2). Therefore, the spring constant can be calculated as follows:

k = (0.67g)(9.8 m/s^2)/(0.096 m)
k = 6.9 N/m

Now, we can use the formula for elastic potential energy to find the energy stored in the stretched band at a displacement of 3.6 m:

Ee = 1/2 kx^2
Ee = 1/2 (6.9 N/m)(3.6 m)^2
Ee = 44.352 J

Therefore, the elastic potential energy stored in the stretched band at a displacement of 3.6 m is 44.352 Joules. This energy is converted into kinetic energy as the elastic band is released, causing it to move with a horizontal velocity. The kinetic energy can then be calculated using the formula Ek = 1/2 mv^2, where m is the mass of the elastic band and v is the velocity. This kinetic energy will continue to decrease as the elastic band slows down due to air resistance and other external factors.

In conclusion, the elastic potential energy stored in a stretched elastic band can be calculated using the formula Ee = 1/2 kx^2, where k is the spring constant and x is the displacement from the equilibrium position. In this case, we found that the elastic potential energy stored
 

Related to Elastic Potential Energy Stored

1. What is elastic potential energy stored?

Elastic potential energy stored is the energy stored in an object due to its deformation, such as stretching or compressing, caused by an external force.

2. What factors affect the amount of elastic potential energy stored?

The amount of elastic potential energy stored depends on the amount of force applied, the distance the object is stretched or compressed, and the material properties of the object, such as its spring constant.

3. How is elastic potential energy stored calculated?

The formula for calculating elastic potential energy stored is PE = 1/2kx^2, where PE is the elastic potential energy, k is the spring constant, and x is the displacement of the object.

4. What are some real-life examples of elastic potential energy stored?

Some examples of elastic potential energy stored include a stretched rubber band, a compressed spring, a bow and arrow, and a trampoline.

5. How is elastic potential energy stored related to kinetic energy?

Elastic potential energy stored and kinetic energy are interrelated. When an object is released from its stretched or compressed state, the elastic potential energy is converted into kinetic energy as the object returns to its original shape.

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