A spring with spring constant 18 N/m has a length of 10 cm. It is pulled to a length of 16 cm. What is the work required to do this?

According to Hooke’s Law, the force required to increase
the length of a spring is proportional to the length by which the spring has been pulled
and is given by the relation as F = kx, where k is the spring constant and x is the
change from the normal length.  In this problem the spring has a spring constant of
18N/m.

When the spring is 10 cm long we require a force of
0 N to pull it. This gradually increases as the spring is pulled further by an
infinitesimal length dL and at 16 cm it is equal to 18*.06
N.

To find the total work done in pulling the spring from
10 cm to 16 cm, we have to find the definite integral of kx for x = 0 to x =
6.

Work required is Int [kx], x= 0 to x =
.06m

=> kx^2/2, x = 0 to x =
0.06

=> 18*x^2/2, x = 0 to x =
0.06

=> 9*x^2, x = 0 to x =
0.06

At x = 0, this is equal to
0.

At x = 0.06, this is equal to 9*0.06^2 =
0.0324

Subtracting the value at x = 0 from that at x = 6
cm, we get 0.0324 J.

Therefore the work
required to pull the spring from 10 cm to 16 cm is
0.0324J.