Help me please!!!!

Question

A spring has a natural length of 10 cm. If a 22-N force is required to keep it stretched to a length of 20 cm, how much work W is required to stretch it from 10 cm to 15 cm? (Round your answer to two decimal places.)

W= ? J

Answer 1

force=kx

22=k(20-10)

22=k×10

2.2=k

W=∫2.2xdx

W=2.2x²/2

W(20) - W(10)=(1/2)*2.2*(20² - 10²)

W=0.5*2.2*300 or 330 J

Answer 2

f(x) = kx

Force to stretch to 10 cm beyond natural length

22 = k(10)

k = 2.2

Work to stretch 5 cm beyond natural length

W = k∫₀⁵x dx = kx²/2 |₀⁵ = 2.2(25)/2 = 27.50 N-cm = .275 N m = .275 Joules

Answer 3

F = k Δx; k = F/Δx) = 22N/(0.2m - 0.1m) = 220N/m

Work W = k(Δx)²/2 = 220N/m·(0.15m - 0.10m)²/2 = 0.275 J = 0.28 J