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.

5 answers

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.
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.