Asked by Favour
A spring of length 2cm requires a force of 6N to stretch it by 0.2cm.
What is the work done in stretching it by 2cm if the elastic limit is not exceeded.
What is the work done in stretching it by 2cm if the elastic limit is not exceeded.
Answers
Answered by
Bot
We can use the formula for the potential energy stored in a spring:
U = (1/2)kx^2
where U is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.
To find the spring constant, we can use the given information:
F = kx
6N = k(0.2cm)
k = 30 N/cm
Now we can find the work done in stretching the spring by 2cm:
U = (1/2)kx^2
U = (1/2)(30 N/cm)(2cm)^2
U = 60 J
So the work done in stretching the spring by 2cm is 60 Joules.
U = (1/2)kx^2
where U is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.
To find the spring constant, we can use the given information:
F = kx
6N = k(0.2cm)
k = 30 N/cm
Now we can find the work done in stretching the spring by 2cm:
U = (1/2)kx^2
U = (1/2)(30 N/cm)(2cm)^2
U = 60 J
So the work done in stretching the spring by 2cm is 60 Joules.
Answered by
Bot
You are correct, we can also use the formulas for elastic potential energy. The elastic potential energy stored in a spring can be calculated using either:
Elastic Potential Energy = 1/2 x spring force x extension (1/2 F x Δx)
or
Elastic Potential Energy = 1/2 x spring constant x extension^2 (1/2 k x^2)
Using the second formula:
The spring constant (k) can be calculated as:
k = F / Δx = 6 N / 0.2 cm = 30 N/cm
The extension (x) when the spring is stretched by 2 cm is:
x = 2 cm - 0 cm = 2 cm
The elastic potential energy stored in the spring can be calculated as:
Elastic Potential Energy = 1/2 k x^2 = 1/2 x 30 N/cm x (2 cm)^2 = 60 J
Therefore, the work done in stretching the spring by 2 cm is 60 Joules, assuming the elastic limit is not exceeded.
Elastic Potential Energy = 1/2 x spring force x extension (1/2 F x Δx)
or
Elastic Potential Energy = 1/2 x spring constant x extension^2 (1/2 k x^2)
Using the second formula:
The spring constant (k) can be calculated as:
k = F / Δx = 6 N / 0.2 cm = 30 N/cm
The extension (x) when the spring is stretched by 2 cm is:
x = 2 cm - 0 cm = 2 cm
The elastic potential energy stored in the spring can be calculated as:
Elastic Potential Energy = 1/2 k x^2 = 1/2 x 30 N/cm x (2 cm)^2 = 60 J
Therefore, the work done in stretching the spring by 2 cm is 60 Joules, assuming the elastic limit is not exceeded.
Answered by
Bot
Sure, here's the complete solution using only the formula for elastic potential energy:
Given:
- Length of the spring (L) = 2 cm
- Spring force required to stretch the spring by 0.2 cm (F) = 6 N
- Extension when the spring is stretched by 2 cm (x) = 2 cm
Formula:
Elastic Potential Energy = 1/2 x spring constant x extension^2 (1/2 k x^2)
where spring constant (k) = F / Δx
Solution:
First, we need to find the spring constant (k) using the given force and extension:
k = F / Δx = 6 N / 0.2 cm = 30 N/cm
Next, we can use the formula for elastic potential energy to find the work done in stretching the spring by 2 cm:
Elastic Potential Energy = 1/2 k x^2
= 1/2 x 30 N/cm x (2 cm)^2
= 60 J
Therefore, the work done in stretching the spring by 2 cm is 60 Joules, assuming the elastic limit is not exceeded.
Given:
- Length of the spring (L) = 2 cm
- Spring force required to stretch the spring by 0.2 cm (F) = 6 N
- Extension when the spring is stretched by 2 cm (x) = 2 cm
Formula:
Elastic Potential Energy = 1/2 x spring constant x extension^2 (1/2 k x^2)
where spring constant (k) = F / Δx
Solution:
First, we need to find the spring constant (k) using the given force and extension:
k = F / Δx = 6 N / 0.2 cm = 30 N/cm
Next, we can use the formula for elastic potential energy to find the work done in stretching the spring by 2 cm:
Elastic Potential Energy = 1/2 k x^2
= 1/2 x 30 N/cm x (2 cm)^2
= 60 J
Therefore, the work done in stretching the spring by 2 cm is 60 Joules, assuming the elastic limit is not exceeded.
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.