Question

Potential energy of certain spring when stretched through a distance of x=10J. The amount of work that must be done on this spring to stretch it through an additional distance x will be

Answer 1

Force due to spring = kx
So,
By integration wrt dx

Work done = F*dx = 1/2*k*x^2

So,
given that,
1/2*k*x^2 = 10 Joule

Now,
For distance of 2x
1/2*k*(2x)^2 = 40 Joule

So,
30 Joules of work has to be performed.