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.
Not correct,use1/2fe or 1/2ke^2 to solve
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.
I want it arrange with the use of formula
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.
To find the work done in stretching the spring by 2cm, we need to use Hooke's Law and the formula for work.
Hooke's Law states that the force required to stretch or compress a spring is directly proportional to the displacement. Mathematically, it can be expressed as:
F = k * x
Where:
F is the force applied to the spring,
k is the spring constant, and
x is the displacement from the equilibrium position.
Let's solve the problem step by step:
Step 1: Calculate the spring constant (k).
Given that a force of 6N is required to stretch the spring by 0.2cm, we can use Hooke's Law to find the spring constant (k).
F = k * x
6N = k * 0.2cm
To convert cm to meters, we divide by 100:
6N = k * 0.002m
Rearranging the equation to solve for k:
k = 6N / 0.002m
k = 3000 N/m
So, the spring constant (k) is 3000 N/m.
Step 2: Calculate the work done in stretching the spring by 2cm.
The work done (W) is given by the formula:
W = (1/2) * k * x^2
Where x is the displacement.
Now, we need to find x when the spring is stretched by 2cm (0.02m) from its original length of 2cm (0.02m).
W = (1/2) * 3000 N/m * (0.02m)^2
W = (1/2) * 3000 N/m * 0.0004m^2
W = 0.6 Nm
W = 0.6 Joules
Therefore, the work done in stretching the spring by 2cm (0.02m) is 0.6 Joules.
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.