this is a question about work done by a variable force.?

Question

A work W0 is required to stretch a certain spring 2 cm from its equilibrium position.
Suppose the spring is already stretched 2 cm from equilibrium. How much additional work (in terms of W0) is required to stretch it to 3 cm from equilibrium?

Answer 1

Let k be the spring constant, and F be the force required to extend it a distance x.

Then:
F = kx

The additional work to stretch is a further small amount dx is:
dw = F dx = kx dx.

Integrating:
w = kx^2 / 2 + const.

As w = 0 when x = 0, the constant is 0:
w = kx^2 / 2

When x = 2cm, w = W0.
Therefore:
W0 = 4k / 2
k = W0 / 2

When x = 3cm:
w = 9k / 2 = 9W0 / 4

The additional work required is:
w - W0
= (9W0 - 4W0) / 4
= 5W0 / 4.

Answer 2

W0= (1/2) k x^2
k - spring constant
x - distance streached
we have

Wd= W1-W0
Wd=(1/2) k( x1^2 -x0^2)
Wd=(1/2) k( (3)^2 -(2)^2)
Wd= (5/2)k