Find the work done (in joules) if the spring is stretched from its natural length to 0.65 meters.

Question

Consider a spring. Recall that the force required to stretch a spring by x units beyond its original

length is given by F(x) = kx, where k is a constant called the spring constant. Suppose a spring is

0.5 meters long, and requires a force of 10 Newtons to stretch it to 0.55 meters. Find the work done

(in joules) if the spring is stretched from its natural length to 0.65 meters.

Answer 1

F = k(x₁ - x₀); k = F/(x₁ - x₀) = 10N/(0.55m - 0.5m) = 200N/m

Work W = ∫₀^0.15200xdx = 100x²₀^0.15 = 100(0.0225 - 0) = 2.25 J

Answer 2

So to find this, we first need to find the spring constant k. We can do this by integrating the general function F(x) from 0.5 to 0.55 and setting this equal to 10. From there, it's rather simple to solve for k.

∫_.5^.55 kx dx = kx²/2 |_.5^.55 = k/2*(.55² - .5²) = .02625k = 10

k ≈ 380.95242

Now from there, you just do the same integration from .5 to .65 now knowing that spring constant k. Since the integral will be the same, I will start where we are plugging in the bounds of the integration.

k/2*(.65² - .5²) = 190.47621(.1725) = 65.7143 N

Feel free to round to wherever you need from there.